question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
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smallest-number-with-all-set-bits | Time Complexity : O( logn ) | Space Complexity : O( 1 ) | time-complexity-o-logn-space-complexity-ifp5f | Code | Trigun_2005 | NORMAL | 2025-01-18T13:55:52.004339+00:00 | 2025-01-18T13:55:52.004339+00:00 | 1 | false | # Code
```cpp []
class Solution {
public:
int smallestNumber(int n) {
int result = 0;
while(n > result)
result = result*2 + 1;
return result;
}
};
``` | 0 | 0 | ['C++'] | 0 |
smallest-number-with-all-set-bits | C Solution | c-solution-by-dmpriya-fh9m | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | dmpriya | NORMAL | 2025-01-17T06:46:59.386003+00:00 | 2025-01-17T06:46:59.386003+00:00 | 17 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['C'] | 0 |
smallest-number-with-all-set-bits | Beats 100% normal while loop approach | beats-100-normal-while-loop-approach-by-yqy94 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | user2633MX | NORMAL | 2025-01-17T06:24:17.310309+00:00 | 2025-01-17T06:24:17.310309+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
smallest-number-with-all-set-bits | scala oneliner | scala-oneliner-by-vititov-o7nc | null | vititov | NORMAL | 2025-01-16T23:05:11.011343+00:00 | 2025-01-16T23:05:11.011343+00:00 | 2 | false | ```scala []
object Solution {
def smallestNumber(n: Int): Int = (1 << BigInt(n).bitLength)-1
}
``` | 0 | 0 | ['Scala'] | 0 |
smallest-number-with-all-set-bits | one liner | one-liner-by-raviteja_29-alow | Intuitionlongest number possible is the length of the bitsApproachchange all 0s in bits to 1Complexity
Time complexity: O(1)
Space complexity: O(1)
Code | raviteja_29 | NORMAL | 2025-01-16T18:06:56.658429+00:00 | 2025-01-16T18:06:56.658429+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
longest number possible is the length of the bits
# Approach
<!-- Describe your approach to solving the problem. -->
change all 0s in bits to 1
# Complexity
- Time complexity: O(1)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- S... | 0 | 0 | ['Math', 'Bit Manipulation', 'Python3'] | 0 |
smallest-number-with-all-set-bits | Easiest soln with 100% beats; | easiest-soln-with-100-beats-by-mamthanag-lb9z | IntuitionApproachTHe idea is simple if n contains a no with all bit set then if we and it with the next no the ans be 0 if its 0 then return n else -1Eg: n=5, | mamthanagaraju1 | NORMAL | 2025-01-16T06:37:16.087934+00:00 | 2025-01-16T06:37:16.087934+00:00 | 13 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
THe idea is simple if n contains a no with all bit set then if we and it with the next no the ans be 0 if its 0 then return n else -1
Eg: n=5,
when n=7 ,n+1=8 so 111&1000... | 0 | 0 | ['C++'] | 0 |
smallest-number-with-all-set-bits | Log(n) easy understanding solution | logn-easy-understanding-solution-by-spid-1sbs | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | spidemen | NORMAL | 2025-01-14T00:17:36.079161+00:00 | 2025-01-14T00:17:36.079161+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Java'] | 0 |
smallest-number-with-all-set-bits | Smallest_number_with_all_set_bits solution in java | smallest_number_with_all_set_bits-soluti-8fug | ApproachUse the property of (&) operator which return 1 if both value is one.So for num with setbits num &(num+1) will be 0
explanation:
7 --> 111
8 -- | abi_shree | NORMAL | 2025-01-12T17:02:41.169044+00:00 | 2025-01-12T17:02:41.169044+00:00 | 3 | false |
# Approach
Use the property of (&) operator which return 1 if both value is one.
So for num with setbits num &(num+1) will be 0
>>explanation:
>>> 7 --> 111
>>> 8 --> 1000
>>> 7 & 8 --> 0
```java []
class Solution {
public int smallestNumber(int n) {
int num=n;
while((num & (nu... | 0 | 0 | ['Java'] | 0 |
smallest-number-with-all-set-bits | 3370 solution in java | 3370-solution-in-java-by-abi_shree-zqpg | ApproachUse the property of (&) operator which return 1 if both value is one.
So for num with setbits num &(num+1) will be 0
explanation:
7 --> 111 | abi_shree | NORMAL | 2025-01-12T16:43:20.530948+00:00 | 2025-01-12T16:43:20.530948+00:00 | 3 | false |
# Approach
Use the property of (&) operator which return 1 if both value is one.
> So for num with setbits num &(num+1) will be 0
>> explanation:
>>> 7 --> 111
>>> 8 --> 1000
>>> 7 & 8 --> 0
# Code
```java []
class Solution {
public int smallestNumber(int n) {
int nu... | 0 | 0 | ['Math', 'Bit Manipulation', 'Java'] | 0 |
populating-next-right-pointers-in-each-node | ✅ [C++/Python/Java] Simple Solution w/ Images & Explanation | BFS + DFS + O(1) Optimized BFS | cpythonjava-simple-solution-w-images-exp-r5nv | We are given a perfect binary tree and we need to populate next pointers in each node of the tree\n\n---\n\n\u2714\uFE0F Solution - I (BFS - Right to Left)\n\nI | archit91 | NORMAL | 2021-12-29T06:23:47.162274+00:00 | 2021-12-29T13:53:54.160080+00:00 | 43,494 | false | We are given a perfect binary tree and we need to populate next pointers in each node of the tree\n\n---\n\n\u2714\uFE0F ***Solution - I (BFS - Right to Left)***\n\nIt\'s important to see that the given tree is a **perfect binary tree**. This means that each node will always have both children and only the last level o... | 1,013 | 2 | [] | 48 |
populating-next-right-pointers-in-each-node | A simple accepted solution | a-simple-accepted-solution-by-ragepyre-73pv | void connect(TreeLinkNode *root) {\n if (root == NULL) return;\n TreeLinkNode *pre = root;\n TreeLinkNode *cur = NULL;\n while(pre-> | ragepyre | NORMAL | 2014-07-09T15:44:45+00:00 | 2018-10-19T08:48:42.927941+00:00 | 148,030 | false | void connect(TreeLinkNode *root) {\n if (root == NULL) return;\n TreeLinkNode *pre = root;\n TreeLinkNode *cur = NULL;\n while(pre->left) {\n cur = pre;\n while(cur) {\n cur->left->next = cur->right;\n if(cur->next) cur->right->next = c... | 1,007 | 11 | [] | 93 |
populating-next-right-pointers-in-each-node | Java solution with O(1) memory+ O(n) time | java-solution-with-o1-memory-on-time-by-t66my | \n\n public class Solution {\n public void connect(TreeLinkNode root) {\n TreeLinkNode level_start=root;\n while(level_start!=nu | talent58 | NORMAL | 2014-12-20T21:11:08+00:00 | 2018-10-26T01:51:20.214018+00:00 | 77,186 | false | \n\n public class Solution {\n public void connect(TreeLinkNode root) {\n TreeLinkNode level_start=root;\n while(level_start!=null){\n TreeLinkNode cur=level_start;\n while(cur!=null){\n if(cur.left!=null) cur.left.next=cur.right;\n ... | 577 | 6 | [] | 50 |
populating-next-right-pointers-in-each-node | 7 lines, iterative, real O(1) space | 7-lines-iterative-real-o1-space-by-stefa-loyz | Simply do it level by level, using the next-pointers of the current level to go through the current level and set the next-pointers of the next level.\n\nI say | stefanpochmann | NORMAL | 2015-06-18T20:50:24+00:00 | 2018-10-22T06:54:24.726050+00:00 | 39,165 | false | Simply do it level by level, using the `next`-pointers of the current level to go through the current level and set the `next`-pointers of the next level.\n\nI say "real" O(1) space because of the many recursive solutions ignoring that recursion management needs space.\n\n def connect(self, root):\n while roo... | 423 | 16 | ['Python'] | 50 |
populating-next-right-pointers-in-each-node | My recursive solution(Java) | my-recursive-solutionjava-by-gnayoaix-xilt | \n public void connect(TreeLinkNode root) {\n if(root == null)\n return;\n \n if(root.left != null){\n root.le | gnayoaix | NORMAL | 2015-04-17T22:09:45+00:00 | 2018-10-23T07:33:20.702260+00:00 | 35,413 | false | \n public void connect(TreeLinkNode root) {\n if(root == null)\n return;\n \n if(root.left != null){\n root.left.next = root.right;\n if(root.next != null)\n root.right.next = root.next.left;\n }\n \n connect(root.left);\n ... | 284 | 2 | [] | 32 |
populating-next-right-pointers-in-each-node | My simple non-iterative C++ code with O(1) memory | my-simple-non-iterative-c-code-with-o1-m-abk3 | void connect(TreeLinkNode *root) {\n if(!root)\n return;\n while(root -> left)\n {\n TreeLinkNode *p = root;\n | erudy | NORMAL | 2015-03-03T09:08:30+00:00 | 2018-10-17T05:27:01.699344+00:00 | 24,423 | false | void connect(TreeLinkNode *root) {\n if(!root)\n return;\n while(root -> left)\n {\n TreeLinkNode *p = root;\n while(p)\n {\n p -> left -> next = p -> right;\n if(p -> next)\n p -> right -> next = p -> ... | 202 | 5 | [] | 17 |
populating-next-right-pointers-in-each-node | C++ EASY TO SOLVE || Beginner Friendly with detailed explanations and dry run | c-easy-to-solve-beginner-friendly-with-d-e84d | Behold the legendary battle between Recursive and Interative Approaches \n\nFight!!!\n\nIntuition:-\n We are given a prefect binary tree that means every parent | Cosmic_Phantom | NORMAL | 2021-12-29T03:27:55.387886+00:00 | 2024-08-23T02:29:09.481253+00:00 | 15,621 | false | > # **Behold the legendary battle between Recursive and Interative Approaches** \n***\n***Fight!!!***\n***\n**Intuition:-**\n* We are given a prefect binary tree that means every parent has two children and all the leaves are on the same level . \n* This question is an superior version of binary level order traversa... | 199 | 11 | ['Recursion', 'C', 'Iterator', 'C++'] | 5 |
populating-next-right-pointers-in-each-node | Python solutions (Recursively, BFS+queue, DFS+stack) | python-solutions-recursively-bfsqueue-df-bcpf | def connect1(self, root):\n if root and root.left and root.right:\n root.left.next = root.right\n if root.next:\n ro | oldcodingfarmer | NORMAL | 2015-07-10T23:25:53+00:00 | 2015-07-10T23:25:53+00:00 | 14,723 | false | def connect1(self, root):\n if root and root.left and root.right:\n root.left.next = root.right\n if root.next:\n root.right.next = root.next.left\n self.connect(root.left)\n self.connect(root.right)\n \n # BFS \n def connect2(self, r... | 150 | 2 | ['Stack', 'Depth-First Search', 'Breadth-First Search', 'Queue', 'Python'] | 17 |
populating-next-right-pointers-in-each-node | Python Solution With Explaintion | python-solution-with-explaintion-by-tyr0-xjwr | I want to share how I come up with this solution with you:\n\nSince we are manipulating tree nodes on the same level, it's easy to come up with\na very standard | tyr034 | NORMAL | 2015-09-21T23:29:40+00:00 | 2018-10-18T10:30:04.965308+00:00 | 15,415 | false | I want to share how I come up with this solution with you:\n\nSince we are manipulating tree nodes on the same level, it's easy to come up with\na very standard BFS solution using queue. But because of next pointer, we actually\ndon't need a queue to store the order of tree nodes at each level, we just use a next\np... | 136 | 4 | ['Python'] | 7 |
populating-next-right-pointers-in-each-node | C++ Iterative/Recursive | c-iterativerecursive-by-jianchao-li-od4v | Recursive\n\nSimilar to a level-order traversal, even you are not allowed to use a queue, the next pointer provides you with a way to move to the next node in t | jianchao-li | NORMAL | 2015-07-11T09:02:00+00:00 | 2015-07-11T09:02:00+00:00 | 12,734 | false | **Recursive**\n\nSimilar to a level-order traversal, even you are not allowed to use a `queue`, the `next` pointer provides you with a way to move to the next node in the same level.\n\n```cpp\nclass Solution {\npublic:\n Node* connect(Node* root) {\n Node *pre = root, *cur;\n while (pre) {\n ... | 131 | 2 | ['Recursion', 'Binary Tree', 'Iterator', 'C++'] | 9 |
populating-next-right-pointers-in-each-node | Java 0ms with visual explanation | java-0ms-with-visual-explanation-by-wils-6bh2 | \n\n\nclass Solution {\n public Node connect(Node root) {\n if(root == null) return null;\n if(root.left != null) root.left.next = root.right;\ | wilsoncursino | NORMAL | 2020-12-07T00:12:47.033875+00:00 | 2020-12-07T00:12:47.033919+00:00 | 6,016 | false | \n\n```\nclass Solution {\n public Node connect(Node root) {\n if(root == null) return null;\n if(root.left != null) root.left.next = root.right;\n if(root.right != null && root.next != ... | 129 | 1 | ['Java'] | 16 |
populating-next-right-pointers-in-each-node | C++ || BFS || Iterative || queue | c-bfs-iterative-queue-by-saurav28-l2nm | \n/*\n// Definition for a Node.\nclass Node {\npublic:\n int val;\n Node* left;\n Node* right;\n Node* next;\n\n Node() : val(0), left(NULL), rig | saurav28_ | NORMAL | 2020-11-13T09:48:39.066397+00:00 | 2020-11-13T09:50:03.951228+00:00 | 5,757 | false | ```\n/*\n// Definition for a Node.\nclass Node {\npublic:\n int val;\n Node* left;\n Node* right;\n Node* next;\n\n Node() : val(0), left(NULL), right(NULL), next(NULL) {}\n\n Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}\n\n Node(int _val, Node* _left, Node* _right, Node* _nex... | 76 | 2 | ['Breadth-First Search', 'Queue', 'C', 'C++'] | 6 |
populating-next-right-pointers-in-each-node | ✅ [Python] Two Solutions || BFS and DFS || Image Explanation || Beginner Friendly | python-two-solutions-bfs-and-dfs-image-e-agwc | PLEASE UPVOTE if you like \uD83D\uDE01 If you have any question, feel free to ask. \n Solution 1\n\t Standard BFS with mantaining pre_level and pre_node as the | linfq | NORMAL | 2021-12-29T05:43:06.330243+00:00 | 2021-12-29T07:56:07.689822+00:00 | 6,355 | false | **PLEASE UPVOTE if you like** \uD83D\uDE01 **If you have any question, feel free to ask.** \n* **Solution 1**\n\t* Standard BFS with mantaining `pre_level` and `pre_node` as the previous node in BFS sequence\n\t\t* `level == pre_level` means current `node` is not the first node of `level`, then `pre_node.next = node` a... | 62 | 1 | ['Depth-First Search', 'Breadth-First Search', 'Python'] | 7 |
populating-next-right-pointers-in-each-node | [Python] O(n) time/ O(log n) space recursion, explained | python-on-time-olog-n-space-recursion-ex-hx69 | In this problem we are given that our tree is perfect binary tree, which will help us a lot. Let us use recursion: imagine, that for left and right subtees we a | dbabichev | NORMAL | 2020-11-13T08:55:18.859593+00:00 | 2020-11-13T08:55:18.859631+00:00 | 2,660 | false | In this problem we are given that our tree is perfect binary tree, which will help us a lot. Let us use recursion: imagine, that for left and right subtees we already make all connections, what we need to connect now? See the image and it will become very clear: we need to connect just `O(log n)` pairs now: we go the t... | 61 | 1 | ['Recursion'] | 4 |
populating-next-right-pointers-in-each-node | Java solution traversing by level without extra space | java-solution-traversing-by-level-withou-k2bu | public class Solution {\n public void connect(TreeLinkNode root) {\n if(root==null) return;\n TreeLinkNode cur = root;\n | upthehell | NORMAL | 2016-05-12T15:07:46+00:00 | 2018-10-17T07:26:09.500360+00:00 | 8,743 | false | public class Solution {\n public void connect(TreeLinkNode root) {\n if(root==null) return;\n TreeLinkNode cur = root;\n TreeLinkNode nextLeftmost = null;\n\n while(cur.left!=null){\n nextLeftmost = cur.left; // save the start of next level\n ... | 56 | 1 | [] | 3 |
populating-next-right-pointers-in-each-node | 💡JavaScript BFS & DFS Solution | javascript-bfs-dfs-solution-by-aminick-4wwo | The Idea - BFS\n1. BFS using queue\n2. as we are shifing node, connect it to the next in queue\njavascript\nvar connectBFS = function(root) {\n if (root == n | aminick | NORMAL | 2019-11-04T11:37:19.860701+00:00 | 2019-11-04T11:37:19.860736+00:00 | 2,928 | false | #### The Idea - BFS\n1. BFS using queue\n2. as we are shifing node, connect it to the next in queue\n``` javascript\nvar connectBFS = function(root) {\n if (root == null) return root;\n let queue = [root];\n while(queue.length!=0) {\n let next = [];\n while(queue.length!=0) {\n let nod... | 44 | 0 | ['JavaScript'] | 5 |
populating-next-right-pointers-in-each-node | BFS || c++ || iterative || explanation || level order traversal | bfs-c-iterative-explanation-level-order-vtl17 | \nBasically this is purely level order travsersal code with slight modification for the root -> next value \n\nYou just have to think 2 things in this question | walkytalkyshubham | NORMAL | 2021-05-06T21:58:00.692069+00:00 | 2022-08-06T16:35:31.773786+00:00 | 2,243 | false | \nBasically this is purely level order travsersal code with slight modification for the root -> next value \n\nYou just have to think 2 things in this question.\n\n1.How to get the last val to NULL ?.\n2.How to get connect with the current node to previous one ?.\n\nIf you are able to find the ans of these two questio... | 40 | 0 | ['Breadth-First Search', 'C', 'C++'] | 4 |
populating-next-right-pointers-in-each-node | [Python] 3 approaches - Clean & Concise | python-3-approaches-clean-concise-by-hie-wg1k | \u2714\uFE0F Solution 1: BFS\npython\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n if root == None: return None\n\n q = d | hiepit | NORMAL | 2020-02-21T14:53:30.000599+00:00 | 2021-09-08T07:00:39.893987+00:00 | 707 | false | **\u2714\uFE0F Solution 1: BFS**\n```python\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n if root == None: return None\n\n q = deque([root])\n while q:\n prev = None\n for _ in range(len(q)):\n curr = q.popleft()\n if prev... | 40 | 0 | [] | 0 |
populating-next-right-pointers-in-each-node | [Python3] BFS and DFS | python3-bfs-and-dfs-by-zhanweiting-edfj | BFS\n\n"""\n# Definition for a Node.\nclass Node:\n def __init__(self, val, left, right, next):\n self.val = val\n self.left = left\n se | zhanweiting | NORMAL | 2019-09-10T07:45:43.353090+00:00 | 2019-09-10T08:08:34.543883+00:00 | 4,333 | false | * BFS\n```\n"""\n# Definition for a Node.\nclass Node:\n def __init__(self, val, left, right, next):\n self.val = val\n self.left = left\n self.right = right\n self.next = next\n"""\nfrom collections import deque\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n ... | 34 | 0 | ['Breadth-First Search', 'Python', 'Python3'] | 6 |
populating-next-right-pointers-in-each-node | Simple recursive Java solution O(1) space O(n) time | simple-recursive-java-solution-o1-space-61bmi | public void connect(TreeLinkNode root) {\n \n if(root==null) return ;\n \n link(root.left,root.right);\n }\n \n //HELPER FU | zihao_li | NORMAL | 2015-11-02T04:19:41+00:00 | 2015-11-02T04:19:41+00:00 | 5,789 | false | public void connect(TreeLinkNode root) {\n \n if(root==null) return ;\n \n link(root.left,root.right);\n }\n \n //HELPER FUNCTION TO LINK TWO NODES TOGETHER\n public void link(TreeLinkNode left, TreeLinkNode right){\n \n if(left==null && right==null) return ;\n ... | 33 | 1 | ['Recursion', 'Java'] | 7 |
populating-next-right-pointers-in-each-node | Java | Step-by-step Explanation | java-step-by-step-explanation-by-sherrie-91m6 | Please upvote if this helps! Thx :D\n\nclass Solution {\n public Node connect(Node root) {\n if (root == null) return null;\n connectTwoNodes(r | SherrieCao | NORMAL | 2021-12-18T14:03:22.347251+00:00 | 2021-12-18T14:03:22.347281+00:00 | 1,923 | false | ## Please upvote if this helps! Thx :D\n```\nclass Solution {\n public Node connect(Node root) {\n if (root == null) return null;\n connectTwoNodes(root.left, root.right);\n return root;\n }\n \n private void connectTwoNodes(Node n1, Node n2){\n if (n1 == null || n2 == null) retu... | 30 | 0 | ['Breadth-First Search', 'Recursion', 'Java'] | 4 |
populating-next-right-pointers-in-each-node | [JAVA] Clean Code, O(N) Time Complexity, 100% Faster Solution | java-clean-code-on-time-complexity-100-f-n0lq | \n/*\n// Definition for a Node.\nclass Node {\n public int val;\n public Node left;\n public Node right;\n public Node next;\n\n public Node() {} | anii_agrawal | NORMAL | 2020-11-13T08:21:16.461725+00:00 | 2020-11-13T08:23:03.828372+00:00 | 1,245 | false | ```\n/*\n// Definition for a Node.\nclass Node {\n public int val;\n public Node left;\n public Node right;\n public Node next;\n\n public Node() {}\n \n public Node(int _val) {\n val = _val;\n }\n\n public Node(int _val, Node _left, Node _right, Node _next) {\n val = _val;\n ... | 30 | 1 | [] | 6 |
populating-next-right-pointers-in-each-node | C++ short recursive solution, no extra space needed | c-short-recursive-solution-no-extra-spac-mmy6 | Key points:\n Use parent\'s next arrow to find right children\'s next buddy in the neighboring tree.\n In short: root->right->next = root->next->left.\n* Take c | lisongsun | NORMAL | 2021-02-11T00:59:03.200888+00:00 | 2021-02-11T01:00:17.995185+00:00 | 1,043 | false | Key points:\n* Use parent\'s next arrow to find right children\'s next buddy in the neighboring tree.\n* In short: root->right->next = root->next->left.\n* Take care of current level\'s children\'s next arrow problem before move down to children subtree.\n```\n Node* connect(Node* root) {\n if (root) {\n if (r... | 29 | 1 | ['Recursion', 'C'] | 4 |
populating-next-right-pointers-in-each-node | 🔥 JavaScript : O(1) space, O(n) time 🔥 | javascript-o1-space-on-time-by-akshaymar-ezyx | We iteratively move from each node to the next node, while fixing the next pointers of their children. \n\nvar connect = function(root) {\n let ptr = root;\n | akshaymarch7 | NORMAL | 2020-07-27T12:54:27.369931+00:00 | 2020-07-30T02:37:25.971688+00:00 | 1,333 | false | We iteratively move from each node to the next node, while fixing the next pointers of their children. \n```\nvar connect = function(root) {\n let ptr = root;\n while(root && root.left){\n let temp = root;\n while(temp) {\n temp.left.next = temp.right;\n temp.right.next = temp.... | 25 | 2 | ['Iterator', 'JavaScript'] | 0 |
populating-next-right-pointers-in-each-node | ✅💯🔥Simple Code🚀📌|| 🔥✔️Easy to understand🎯 || 🎓🧠Beats 100%🔥|| Beginner friendly💀💯 | simple-code-easy-to-understand-beats-100-qta0 | Tuntun Mosi ko Pranam\nSolution tuntun mosi ki photo ke baad hai. Scroll Down\n\n# Code\njava []\n/*\n// Definition for a Node.\nclass Node {\n public int va | atishayj4in | NORMAL | 2024-09-10T18:14:34.415993+00:00 | 2024-09-10T18:14:34.416029+00:00 | 1,663 | false | # Tuntun Mosi ko Pranam\nSolution tuntun mosi ki photo ke baad hai. Scroll Down\n\n# Code\n```java []\n/*\n// Definition for a Node.\nclass Node {\n publi... | 23 | 1 | ['Linked List', 'Tree', 'Depth-First Search', 'Breadth-First Search', 'C', 'Binary Tree', 'Python', 'C++', 'Java', 'JavaScript'] | 2 |
populating-next-right-pointers-in-each-node | 5 lines C++ simple solution. | 5-lines-c-simple-solution-by-tiny656-t8i9 | class Solution {\n public:\n void connect(TreeLinkNode *root) {\n if (!root) return;\n if (root->left) root->left->next = root-> | tiny656 | NORMAL | 2015-08-09T09:39:30+00:00 | 2015-08-09T09:39:30+00:00 | 2,064 | false | class Solution {\n public:\n void connect(TreeLinkNode *root) {\n if (!root) return;\n if (root->left) root->left->next = root->right;\n if (root->right && root->next) root->right->next = root->next->left;\n connect(root->left);\n connect(root->right)... | 23 | 1 | [] | 0 |
populating-next-right-pointers-in-each-node | Java Solution | Simple BFS Traversal | Very easy iterative solution | java-solution-simple-bfs-traversal-very-zjbnj | \nclass Solution {\n\n public Node connect(Node root) {\n if (root == null) return root;\n Queue<Node> q = new LinkedList<>();\n q.add(r | 1605448777 | NORMAL | 2022-06-24T12:39:04.345875+00:00 | 2022-06-24T12:39:04.345917+00:00 | 1,098 | false | ```\nclass Solution {\n\n public Node connect(Node root) {\n if (root == null) return root;\n Queue<Node> q = new LinkedList<>();\n q.add(root);\n while (!q.isEmpty()) {\n int size = q.size();\n for (int i = 0; i < size; i++) {\n Node curr = q.poll();\... | 20 | 0 | ['Breadth-First Search', 'Queue', 'Java'] | 1 |
populating-next-right-pointers-in-each-node | Another simple JavaScript solution | another-simple-javascript-solution-by-je-hdsd | Since it's a full binary tree, our job is much simpler, at each node, connects its left and right child, and try to connect the right child with the left child | jeantimex | NORMAL | 2017-10-15T18:34:36.123000+00:00 | 2018-09-13T19:53:39.397636+00:00 | 1,574 | false | Since it's a full binary tree, our job is much simpler, at each node, connects its left and right child, and try to connect the right child with the left child of node's next. A simple preorder traversal should be able to help us solve this problem.\n```\n/**\n * @param {TreeLinkNode} root\n * @return {void} Do not ret... | 19 | 1 | [] | 1 |
populating-next-right-pointers-in-each-node | Very Easy to understand recursive Method | very-easy-to-understand-recursive-method-8gpm | Simple recursive solution accepted\n\nclass Solution {\npublic:\n \n void solve(Node* l, Node* r){\n \n if(l == NULL || r == NULL) return;\n | rachit7399 | NORMAL | 2020-07-21T16:34:54.296970+00:00 | 2020-07-21T16:34:54.297005+00:00 | 913 | false | Simple recursive solution accepted\n```\nclass Solution {\npublic:\n \n void solve(Node* l, Node* r){\n \n if(l == NULL || r == NULL) return;\n \n l->next = r;\n r->next = NULL;\n \n solve(l->left, l->right);\n solve(l->right, r->left);\n solve(r->lef... | 17 | 0 | ['Recursion', 'C', 'C++'] | 4 |
populating-next-right-pointers-in-each-node | python solution | python-solution-by-pankit-1r7n | \nclass Solution:\n # @param root, a tree link node\n # @return nothing\n def helper(self, left, right):\n if not left or not right:\n | pankit | NORMAL | 2018-07-24T02:42:04.679730+00:00 | 2018-10-11T01:56:10.848397+00:00 | 1,364 | false | ```\nclass Solution:\n # @param root, a tree link node\n # @return nothing\n def helper(self, left, right):\n if not left or not right:\n return\n \n left.next = right\n self.helper(left.right, right.left)\n self.helper(left.left, left.right)\n self.helper(r... | 16 | 0 | [] | 2 |
populating-next-right-pointers-in-each-node | Python accepted code | python-accepted-code-by-yasheen-1mv5 | def connect(self, root):\n if not root: return\n while root.left:\n cur = root.left\n prev = None\n while root:\n | yasheen | NORMAL | 2015-08-07T04:18:37+00:00 | 2018-09-18T07:13:08.454507+00:00 | 4,084 | false | def connect(self, root):\n if not root: return\n while root.left:\n cur = root.left\n prev = None\n while root:\n if prev: prev.next = root.left\n root.left.next = root.right\n prev = root.right\n root = root.... | 16 | 0 | [] | 1 |
populating-next-right-pointers-in-each-node | C++ recursive solution | c-recursive-solution-by-cychung-gyxa | \n void connect(TreeLinkNode *root) {\n if(!root) return;\n if(root->left){\n root->left->next = root->right;\n root->rig | cychung | NORMAL | 2017-12-23T05:43:53.078000+00:00 | 2017-12-23T05:43:53.078000+00:00 | 1,003 | false | ```\n void connect(TreeLinkNode *root) {\n if(!root) return;\n if(root->left){\n root->left->next = root->right;\n root->right->next = root->next? root->next->left : NULL;\n }\n connect(root->left);\n connect(root->right);\n }\n``` | 16 | 0 | [] | 2 |
populating-next-right-pointers-in-each-node | Python Solution O(1) and O(n) memory. | python-solution-o1-and-on-memory-by-dars-kofr | \nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n # edge case check\n if not root:\n return None\n \n | darshan_22 | NORMAL | 2020-07-04T15:27:12.675170+00:00 | 2020-07-04T15:27:12.675222+00:00 | 1,663 | false | ```\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n # edge case check\n if not root:\n return None\n \n # initialize the queue with root node (for level order traversal)\n queue = collections.deque([root])\n \n # start the traversal\n ... | 15 | 0 | ['Breadth-First Search', 'Queue', 'Python', 'Python3'] | 2 |
populating-next-right-pointers-in-each-node | JavaScript BFS | javascript-bfs-by-deleted_user-j54q | \nvar connect = function(root) {\n if(!root) return;\n const queue = [root];\n \n while(queue.length) {\n const size = queue.length;\n | deleted_user | NORMAL | 2018-08-03T20:34:17.987521+00:00 | 2018-09-25T23:27:38.217316+00:00 | 1,361 | false | ```\nvar connect = function(root) {\n if(!root) return;\n const queue = [root];\n \n while(queue.length) {\n const size = queue.length;\n const level = queue.slice();\n\n for(let i = 0; i < size; i++) {\n const currentNode = queue.shift();\n currentNode.next = le... | 15 | 2 | [] | 2 |
populating-next-right-pointers-in-each-node | An iterative java solution | an-iterative-java-solution-by-graceluli-nue7 | public void connect(TreeLinkNode root) {\n \n TreeLinkNode n = root;\n \n while(n != null && n.left != null) {\n TreeLink | graceluli | NORMAL | 2015-11-13T22:57:50+00:00 | 2015-11-13T22:57:50+00:00 | 2,361 | false | public void connect(TreeLinkNode root) {\n \n TreeLinkNode n = root;\n \n while(n != null && n.left != null) {\n TreeLinkNode pre = null;\n \n for(TreeLinkNode p = n; p != null; p = p.next) {\n if(pre != null) pre.next = p.left;\n ... | 14 | 0 | [] | 0 |
populating-next-right-pointers-in-each-node | Python Solution - Recursive Elegant Solution | python-solution-recursive-elegant-soluti-4f3p | I thought that this solution was a little different to the others posted, most of them doing a level order search using the next pointer. However here, I have r | wallahwallah | NORMAL | 2021-12-29T02:56:05.235556+00:00 | 2021-12-30T23:56:17.350705+00:00 | 718 | false | I thought that this solution was a little different to the others posted, most of them doing a level order search using the next pointer. However here, I have recursively split the tree into \'pincer\' segments (this is what I call them because I don\'t know the name), at each level the pair is made up of 1. the right ... | 13 | 0 | ['Depth-First Search', 'Recursion', 'Python'] | 2 |
populating-next-right-pointers-in-each-node | javascript DFS extremely simple and understandable | javascript-dfs-extremely-simple-and-unde-m5kc | DFS, pass down rightnode\'s left pointer if it exists, otherwise null.\n\nvar connect = function(root, rightNode = null) {\n if (!root) return root;\n \n | anthonysgro1995 | NORMAL | 2021-09-23T04:12:49.659010+00:00 | 2021-09-24T13:54:35.249292+00:00 | 677 | false | DFS, pass down rightnode\'s left pointer if it exists, otherwise null.\n```\nvar connect = function(root, rightNode = null) {\n if (!root) return root;\n \n root.next = rightNode;\n connect(root.left, root.right);\n connect(root.right, rightNode ? rightNode.left : null);\n \n return root;\n};\n``` | 13 | 0 | ['Depth-First Search', 'JavaScript'] | 1 |
populating-next-right-pointers-in-each-node | C++ | [99%, 100% memory] | 5-liner | Recursively crispy AF | c-99-100-memory-5-liner-recursively-cris-zyhk | \nclass Solution {\npublic:\n Node* connect(Node* root) {\n if(root and root->left) {\n root->left->next = root->right;\n auto c | s4chin | NORMAL | 2020-11-13T08:42:40.149836+00:00 | 2020-11-13T08:44:01.865598+00:00 | 592 | false | ```\nclass Solution {\npublic:\n Node* connect(Node* root) {\n if(root and root->left) {\n root->left->next = root->right;\n auto c1 = root->left, c2 = root->right;\n while(c1->right) c1->right->next = c2->left, c1 = c1->right, c2 = c2->left;\n root->left = connect(... | 13 | 3 | [] | 4 |
populating-next-right-pointers-in-each-node | Accepted Java recursive solution | accepted-java-recursive-solution-by-jean-50hb | The recursive solution of my last post, although the space is not O(1) (due to recursion), the solution is still elegant.\n\n public class Solution {\n | jeantimex | NORMAL | 2015-07-04T18:26:09+00:00 | 2015-07-04T18:26:09+00:00 | 1,870 | false | The recursive solution of my last post, although the space is not O(1) (due to recursion), the solution is still elegant.\n\n public class Solution {\n public void connect(TreeLinkNode root) {\n if (root == null) return;\n \n if (root.left != null) {\n root.left... | 13 | 1 | ['Java'] | 1 |
populating-next-right-pointers-in-each-node | 5-line 1ms java iterative solution O(n) time O(1) space | 5-line-1ms-java-iterative-solution-on-ti-lxcd | public class Solution {\n public void connect(TreeLinkNode root) {\n if (root == null) { return; }\n for (TreeLinkNode head=root; h | mach7 | NORMAL | 2016-01-19T03:56:35+00:00 | 2016-01-19T03:56:35+00:00 | 2,306 | false | public class Solution {\n public void connect(TreeLinkNode root) {\n if (root == null) { return; }\n for (TreeLinkNode head=root; head.left!=null; head=head.left) {\n for (TreeLinkNode parent=head; parent!=null; parent=parent.next) {\n parent.left.next ... | 13 | 0 | ['Iterator', 'Java'] | 2 |
populating-next-right-pointers-in-each-node | O(1) space , simple bfs java solution and without recursion | o1-space-simple-bfs-java-solution-and-wi-7kry | Intution: Treating level of tree is as linked list.\n1. From parent level connect children level node as linked list and parent level is already connected so we | deepakkdkk | NORMAL | 2023-06-09T09:47:53.131196+00:00 | 2023-06-09T09:47:53.131234+00:00 | 1,786 | false | Intution: Treating level of tree is as linked list.\n1. From parent level connect children level node as linked list and parent level is already connected so we can move to next node of parent to connect other children node.\n```\nclass Solution {\n public Node connect(Node root) {\n Node head = root;\n ... | 12 | 0 | ['Breadth-First Search', 'Java'] | 3 |
populating-next-right-pointers-in-each-node | ✔️ 100% Fastest Swift Solution, time: O(n), space: O(1). | 100-fastest-swift-solution-time-on-space-vp8g | \n/**\n * Definition for a Node.\n * public class Node {\n * public var val: Int\n * public var left: Node?\n * public var right: Node?\n *\t publ | sergeyleschev | NORMAL | 2022-04-10T07:08:06.731773+00:00 | 2022-04-10T07:08:06.731805+00:00 | 349 | false | ```\n/**\n * Definition for a Node.\n * public class Node {\n * public var val: Int\n * public var left: Node?\n * public var right: Node?\n *\t public var next: Node?\n * public init(_ val: Int) {\n * self.val = val\n * self.left = nil\n * self.right = nil\n * self.nex... | 12 | 0 | ['Swift'] | 0 |
populating-next-right-pointers-in-each-node | A concise O(1) space complexity solution | a-concise-o1-space-complexity-solution-b-tqf5 | it fits problem 1 and 2, any comments will be welcome, thanks\n\n void connect(TreeLinkNode root) {\n TreeLinkNode head = root; // the left first node | shichaotan | NORMAL | 2015-01-11T22:17:16+00:00 | 2015-01-11T22:17:16+00:00 | 1,542 | false | it fits problem 1 and 2, any comments will be welcome, thanks\n\n void connect(TreeLinkNode *root) {\n TreeLinkNode *head = root; // the left first node in every level\n TreeLinkNode *cur = NULL; // the current node in the upper level\n TreeLinkNode *pre = NULL; // the prev node in the downer ... | 12 | 0 | [] | 0 |
populating-next-right-pointers-in-each-node | Java recursive and iterative solutions. | java-recursive-and-iterative-solutions-b-bfl7 | \n // dfs iteratively \n public void connect1(TreeLinkNode root) {\n Stack<TreeLinkNode> stack = new Stack<>();\n stack.push(root);\n | oldcodingfarmer | NORMAL | 2016-04-24T16:26:26+00:00 | 2016-04-24T16:26:26+00:00 | 1,897 | false | \n // dfs iteratively \n public void connect1(TreeLinkNode root) {\n Stack<TreeLinkNode> stack = new Stack<>();\n stack.push(root);\n while (!stack.isEmpty()) {\n TreeLinkNode n = stack.pop();\n if (n != null) {\n if (n.right != null) {\n ... | 12 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Recursion', 'Iterator', 'Java'] | 2 |
populating-next-right-pointers-in-each-node | O(n)time/BEATS 99.97% MEMORY/SPEED 0ms MAY 2022 | ontimebeats-9997-memoryspeed-0ms-may-202-666a | \n\n\n(Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful, please upvote this post.)\nTake care broth | darian-catalin-cucer | NORMAL | 2022-05-15T04:42:54.079149+00:00 | 2022-05-15T04:42:54.079187+00:00 | 1,955 | false | ```\n```\n\n(Note: This is part of a series of Leetcode solution explanations. If you like this solution or find it useful, ***please upvote*** this post.)\n***Take care brother, peace, love!***\n\n```\n```\n\n```\n```\n\n```\n```\n\nThe best result for the code below is ***0ms / 38.2MB*** (beats 92.04% / 24.00%).\n* *... | 11 | 0 | ['Swift', 'C', 'Python', 'C++', 'Java', 'Python3', 'Kotlin', 'JavaScript'] | 0 |
populating-next-right-pointers-in-each-node | ✅ [C++/Java/Python] | BFS | O(n) Time & O(1) Space | cjavapython-bfs-on-time-o1-space-by-dami-f0y3 | I hope the comments are explicit enough to tell about the iterative BFS approach that is used here.\nTime complexity - O(n)\nSpace complexity - O(1)\n\nC++ \n\n | damian_arado | NORMAL | 2022-04-16T17:08:47.380671+00:00 | 2022-08-11T06:25:19.287605+00:00 | 572 | false | I hope the comments are explicit enough to tell about the iterative BFS approach that is used here.\nTime complexity - O(n)\nSpace complexity - O(1)\n\nC++ \n\n```\nclass Solution {\npublic:\n Node* connect(Node* root) {\n if(!root) return root;\n Node* current = root;\n while(current) {\n ... | 11 | 0 | ['Breadth-First Search', 'C', 'Iterator', 'Java'] | 0 |
populating-next-right-pointers-in-each-node | 🔥BEATS 💯 % 🎯 |✨SUPER EASY BEGINNERS 👏 | beats-super-easy-beginners-by-codewithsp-0jdv | IntuitionThe problem involves connecting all nodes at the same level in a perfect binary tree. Since the tree is perfect (all levels are completely filled), eac | CodeWithSparsh | NORMAL | 2025-01-12T17:08:16.488879+00:00 | 2025-01-12T17:08:16.488879+00:00 | 1,202 | false | 
---
# **Intuition**
The problem involves connecting all nodes at the same level in a perfect binary tree. Since the tree is perfect (all levels are completely filled), each node will have its left chi... | 10 | 0 | ['Breadth-First Search', 'C', 'Binary Tree', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'Dart'] | 2 |
populating-next-right-pointers-in-each-node | O(n) time and O(1) space Easy Intutive solution for better understanding of recursion | on-time-and-o1-space-easy-intutive-solut-dihl | Intuition\n Describe your first thoughts on how to solve this problem. \nInitially, I considered a bottom-up recursion approach where I would set the next point | Rya-man | NORMAL | 2024-08-27T18:29:16.994985+00:00 | 2024-08-27T18:31:26.949912+00:00 | 1,373 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nInitially, I considered a bottom-up recursion approach where I would set the next pointer during the backtracking step. However, this approach failed when connecting a node on the right to another node on its parent\'s left. This led me t... | 10 | 0 | ['Linked List', 'Tree', 'C++', 'Java', 'Python3'] | 0 |
populating-next-right-pointers-in-each-node | 2 ms | Easy to Understand | Using Queue | Level Order Traversal | Java | C++ | 2-ms-easy-to-understand-using-queue-leve-c1fj | Intuition\n Describe your first thoughts on how to solve this problem. \nThe approach used here is based on level-order traversal of the tree using a queue. The | yshivhare163 | NORMAL | 2023-06-27T04:26:37.484586+00:00 | 2023-06-27T04:26:37.484616+00:00 | 632 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe approach used here is based on level-order traversal of the tree using a queue. The intuition behind the solution is that by traversing the tree level by level, we can keep track of the next right node for each node in the current lev... | 10 | 0 | ['Breadth-First Search', 'Queue', 'Binary Tree', 'C++', 'Java'] | 4 |
populating-next-right-pointers-in-each-node | 0ms || 100% FASTER || JAVA CODE | 0ms-100-faster-java-code-by-raghavdabra-8dtc | \nclass Solution {\n public Node connect(Node root) {\n if(root == null) return null;\n Node left = root.left;\n Node right = root.right | raghavdabra | NORMAL | 2022-10-19T05:26:07.222974+00:00 | 2022-10-19T05:26:07.223017+00:00 | 645 | false | ```\nclass Solution {\n public Node connect(Node root) {\n if(root == null) return null;\n Node left = root.left;\n Node right = root.right;\n while(left != null){\n left.next = right;\n left = left.right;\n right = right.left;\n }\n connect(... | 10 | 0 | ['Java'] | 0 |
populating-next-right-pointers-in-each-node | Share my LOOP JAVA 1MS solution!! Easy understand!! | share-my-loop-java-1ms-solution-easy-und-entl | //Just remember to use result from the last step\n public class Solution {\n public void connect(TreeLinkNode root) {\n if(root==null) retu | herrji | NORMAL | 2016-02-20T08:05:13+00:00 | 2016-02-20T08:05:13+00:00 | 1,709 | false | //Just remember to use result from the last step\n public class Solution {\n public void connect(TreeLinkNode root) {\n if(root==null) return;\n while(root.left!=null){\n TreeLinkNode tmp = root;\n while(tmp!=null){\n tmp.left.next = t... | 10 | 2 | ['Java'] | 0 |
populating-next-right-pointers-in-each-node | 📌 C++ || Beats - 97% || Explained Using Level Order Traversal | c-beats-97-explained-using-level-order-t-ocap | Best Solution for Beginners\nWe can solve this using Level order Traversal.\n\n Runtime - 97.14%\uD83D\uDD25 \n\n# Approach\n1. Firstly We will do Level Orde | Luvchaudhary | NORMAL | 2022-12-27T05:22:46.111696+00:00 | 2022-12-27T05:22:46.111739+00:00 | 1,451 | false | # Best Solution for Beginners\nWe can solve this using Level order Traversal.\n\n Runtime - 97.14%\uD83D\uDD25 \n\n# Approach\n1. Firstly We will do Level Order Travesal.\n2. Then at Each Level we will join all nodes present at that Level.\n\n# If you like the solution and understand it then Please Upvote.\u2B06\uF... | 9 | 0 | ['Linked List', 'Tree', 'Breadth-First Search', 'C++'] | 4 |
populating-next-right-pointers-in-each-node | C++ solutions | c-solutions-by-infox_92-fhgr | \nclass Solution {\npublic:\n Node* connect(Node* root) {\n queue<Node*> q;\n if (root) q.push(root);\n while (q.size()) {\n | Infox_92 | NORMAL | 2022-11-05T03:49:57.942902+00:00 | 2022-11-05T03:49:57.942941+00:00 | 1,112 | false | ```\nclass Solution {\npublic:\n Node* connect(Node* root) {\n queue<Node*> q;\n if (root) q.push(root);\n while (q.size()) {\n int len = q.size();\n Node* curr;\n while (len--) {\n curr = q.front(), q.pop();\n curr->next = len ? q.f... | 9 | 0 | ['C', 'C++'] | 0 |
populating-next-right-pointers-in-each-node | Java BFS and DFS - 4 Solutions | java-bfs-and-dfs-4-solutions-by-aakashde-bv38 | \n\t//DFS\n\t//Time Complexity : O(n), where n is the number of elements in root\n\t//Space Complexity : O(log n), for recursion stack of a perfect BST\n\tpubli | aakashdeo | NORMAL | 2022-02-12T07:33:21.325584+00:00 | 2022-02-12T07:33:59.804504+00:00 | 594 | false | ```\n\t//DFS\n\t//Time Complexity : O(n), where n is the number of elements in root\n\t//Space Complexity : O(log n), for recursion stack of a perfect BST\n\tpublic Node connect(Node root) {\n if(root == null)\n return root;\n dfs(root);\n return root;\n }\n \n private void dfs(... | 9 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Queue', 'Java'] | 0 |
populating-next-right-pointers-in-each-node | Python Simple Solution easy to understand | O(n) and constant Memory | python-simple-solution-easy-to-understan-56xc | \n def helper(self, root, parent=None, isleftChild=True):\n if root is None:\n return None\n \n if parent == None:\n | sathwickreddy | NORMAL | 2021-07-23T16:20:09.368414+00:00 | 2021-08-04T11:38:50.117641+00:00 | 880 | false | ```\n def helper(self, root, parent=None, isleftChild=True):\n if root is None:\n return None\n \n if parent == None:\n #we are at root\n root.next = None\n else:\n #we are at some node other than root\n if isleftChild:\n ... | 9 | 0 | ['Python', 'C++', 'Java', 'Python3'] | 0 |
populating-next-right-pointers-in-each-node | easy BFS Python 🐍 solution | easy-bfs-python-solution-by-injysarhan-lwzn | \n\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n \n if not root:\n return None\n \n q=collection | injysarhan | NORMAL | 2020-11-14T05:55:02.859691+00:00 | 2020-11-14T05:55:02.859723+00:00 | 925 | false | ```\n\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n \n if not root:\n return None\n \n q=collections.deque()\n \n q.append(root)\n \n while q:\n for i in range(1,len(q)):\n q[i-1].next=q[i]\n ... | 9 | 0 | ['Python', 'Python3'] | 0 |
populating-next-right-pointers-in-each-node | Populating Next Right Pointer in Each Node | C++ | Easy to Understand | populating-next-right-pointer-in-each-no-cqzq | \nclass Solution {\n\npublic:\n Node* connect(Node* root) {\n if(root == NULL || root->left == NULL) return root;\n \n \n if(root | caffeinatedcod3r | NORMAL | 2020-11-13T14:58:38.578370+00:00 | 2020-11-13T15:19:52.572370+00:00 | 323 | false | ```\nclass Solution {\n\npublic:\n Node* connect(Node* root) {\n if(root == NULL || root->left == NULL) return root;\n \n \n if(root->next != NULL){\n root->right->next=root->next->left;\n }\n \n root->left->next = root->right;\n \n connect(ro... | 9 | 0 | ['Recursion', 'C'] | 1 |
populating-next-right-pointers-in-each-node | [Python] Time O(n), Space O(1), Concise Real O(1) Solution | python-time-on-space-o1-concise-real-o1-kqjuz | \nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n if (head:=root):\n while (start:=root.left):\n while ro | ztonege | NORMAL | 2020-11-13T11:25:15.508293+00:00 | 2020-11-13T11:25:15.508323+00:00 | 277 | false | ```\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n if (head:=root):\n while (start:=root.left):\n while root:\n root.left.next = root.right\n root.right.next = root.next.left if root.next else None\n root =... | 9 | 0 | [] | 0 |
populating-next-right-pointers-in-each-node | Python, constant space | python-constant-space-by-warmr0bot-6dg2 | Use the already-populated next pointers of the row above to fill the row below:\n\ndef connect(self, root: \'Node\') -> \'Node\':\n\tif not root: return\n\t\n\t | warmr0bot | NORMAL | 2020-11-13T10:06:46.898732+00:00 | 2020-11-13T10:07:03.274431+00:00 | 566 | false | Use the already-populated next pointers of the row above to fill the row below:\n```\ndef connect(self, root: \'Node\') -> \'Node\':\n\tif not root: return\n\t\n\tabove, below = root, root.left\n\twhile below:\n\t\tcur = below\n\t\twhile above:\n\t\t\tif cur == above.left:\n\t\t\t\tcur.next = above.right\n\t\t\t\tabove... | 9 | 0 | ['Python', 'Python3'] | 0 |
populating-next-right-pointers-in-each-node | Go Recursive and Iterative | go-recursive-and-iterative-by-casd82-b4x2 | Recursive:\n\nfunc connect(root *Node) *Node {\n if root == nil {\n return nil\n }\n \n if root.Left != nil {\n root.Left.Next = root. | casd82 | NORMAL | 2020-04-03T05:47:37.109853+00:00 | 2020-04-03T05:55:27.866868+00:00 | 642 | false | Recursive:\n```\nfunc connect(root *Node) *Node {\n if root == nil {\n return nil\n }\n \n if root.Left != nil {\n root.Left.Next = root.Right\n if root.Next != nil {\n root.Right.Next = root.Next.Left\n }\n }\n \n connect(root.Left)\n connect(root.Right)\n... | 9 | 0 | ['Go'] | 0 |
populating-next-right-pointers-in-each-node | C++ easy recursive solution | c-easy-recursive-solution-by-jtimberlake-enbl | class Solution {\n public:\n void connect(TreeLinkNode *root) {\n if(!root)\n return;\n if(root->left)\n | jtimberlakers | NORMAL | 2015-08-21T03:08:16+00:00 | 2015-08-21T03:08:16+00:00 | 1,031 | false | class Solution {\n public:\n void connect(TreeLinkNode *root) {\n if(!root)\n return;\n if(root->left)\n root->left->next = root->right;\n if(root->next && root->right)\n root->right->next = root->next->left;\n connec... | 9 | 1 | [] | 2 |
populating-next-right-pointers-in-each-node | Populating Next Right Pointers in Each Node – BFS Solution in C++ 🚀 | populating-next-right-pointers-in-each-n-clqa | Intuition
The goal is to connect each node to its next right node in a perfect binary tree. If there’s no next right node, it should remain NULL.
This is a clas | Mirin_Mano_M | NORMAL | 2025-02-17T10:27:23.997378+00:00 | 2025-02-17T10:27:23.997378+00:00 | 755 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
1) The goal is to connect each node to its next right node in a perfect binary tree. If there’s no next right node, it should remain NULL.
2) This is a classic level-order traversal problem, where we process each level of the tree and conne... | 8 | 0 | ['Linked List', 'Tree', 'Breadth-First Search', 'Binary Tree', 'C++', 'Java'] | 0 |
populating-next-right-pointers-in-each-node | Populating Next Right Pointers in Each Node [C++] | populating-next-right-pointers-in-each-n-7b08 | IntuitionThe problem requires us to populate the next pointers in each node of a perfect binary tree. Given the properties of a perfect binary tree, each node h | moveeeax | NORMAL | 2025-01-29T14:24:55.749612+00:00 | 2025-01-29T14:24:55.749612+00:00 | 591 | false | ### Intuition
The problem requires us to populate the `next` pointers in each node of a **perfect binary tree**. Given the properties of a perfect binary tree, each node has either 0 or 2 children, and all leaves are at the same level.
Since each level is **fully populated**, we can use the `next` pointer to traverse ... | 8 | 0 | ['C++'] | 0 |
populating-next-right-pointers-in-each-node | just 4 lines | very easy for beginners | beats 100% | just-4-lines-very-easy-for-beginners-bea-3pjv | IntuitionIn a perfect binary tree:
All levels are completely filled.
Every parent node has exactly two children.
The goal is to populate the next pointers of ea | sharqawycs | NORMAL | 2025-01-26T15:14:23.689071+00:00 | 2025-01-26T15:14:23.689071+00:00 | 648 | false | # Intuition
In a perfect binary tree:
- All levels are completely filled.
- Every parent node has exactly two children.
The goal is to populate the `next` pointers of each node to point to its immediate right neighbor. If there is no right neighbor, the `next` pointer should point to `null`.
We can take advantage o... | 8 | 0 | ['Tree', 'Recursion', 'Binary Tree', 'Java'] | 3 |
populating-next-right-pointers-in-each-node | 🍀Easy BFS 🔥 | | Java 🔥| | Python 🔥| | C++ 🔥 | easy-bfs-java-python-c-by-niketh_1234-u73r | * Extra space but clean code\n---\njava []\nclass Solution {\n public Node connect(Node root) {\n if(root == null)\n return root;\n | niketh_1234 | NORMAL | 2023-06-27T13:00:32.084933+00:00 | 2023-06-27T13:00:32.084966+00:00 | 1,267 | false | # * Extra space but clean code\n---\n```java []\nclass Solution {\n public Node connect(Node root) {\n if(root == null)\n return root;\n Queue<Node> queue = new LinkedList<>();\n queue.add(root);\n while(queue.size() > 0)\n {\n Deque<Node> dq = new ArrayDeque<... | 8 | 0 | ['Breadth-First Search', 'Python', 'C++', 'Java', 'Python3'] | 0 |
populating-next-right-pointers-in-each-node | ✅C++ || 2 Methods With Logic || BFS Traversal || CLEAN CODE | c-2-methods-with-logic-bfs-traversal-cle-l6mo | Method -1 [Naive Method]\n\n\n\nn==Number of Nodes \nT->O(n) && S->O(n) [For storing all the Nodes after BFS] + O(n/2) [For queue, worst Case as it\'s a Perfect | abhinav_0107 | NORMAL | 2023-01-03T07:38:58.560101+00:00 | 2023-01-03T14:27:29.513254+00:00 | 1,257 | false | ***Method -1 [Naive Method]***\n\n\n\n**n==Number of Nodes \nT->O(n) && S->O(n) [For storing all the Nodes after BFS] + O(n/2) [For queue, worst Case as it\'s a Perfect Binary Tree]**\n\n***Logic -> Level Order... | 8 | 0 | ['Breadth-First Search', 'C', 'C++'] | 1 |
populating-next-right-pointers-in-each-node | JAVA| 100% FAST | 0ms | EASY | RECURSION | NODE | java-100-fast-0ms-easy-recursion-node-by-ucay | If you find my solution helpful please upvote it, as it motivates me to post such kind of codes and help the coding community,\nif you have any queries or any i | Narendra_Madireddy | NORMAL | 2022-09-15T00:59:57.243888+00:00 | 2022-09-15T00:59:57.243930+00:00 | 499 | false | If you find my solution helpful please upvote it, as it motivates me to post such kind of codes and help the coding community,\nif you have any queries or any improvements please feel free to comment and share your views.\n\n class Solution {\n public Node connect(Node root) {\n if(root == null){\n ... | 8 | 0 | ['Recursion', 'Java'] | 0 |
populating-next-right-pointers-in-each-node | Python with simple recursion + explanation with drawing | python-with-simple-recursion-explanation-gyiq | I divide the code into four part, check out the hashtag. \n#part1: if root is None, we want to just return None.\n#part2: : if cur.left is not None, we want to | dingchiun | NORMAL | 2022-09-05T11:53:11.687600+00:00 | 2022-09-18T07:22:31.346886+00:00 | 462 | false | I divide the code into four part, check out the hashtag. \n`#part1`: if root is None, we want to just return None.\n`#part2:` : if cur.left is not None, we want to build a connection between cur.left and cur.right. See pic:\n\n##### Recursive Approach :-\n Base case: if the root is null than return null\n N | Debajyoti-Shit | NORMAL | 2022-02-16T05:37:38.055150+00:00 | 2022-02-16T05:37:38.055179+00:00 | 519 | false | ##### If you understand the approach please please upvote!!!\uD83D\uDC4D\n***Thanks :)***\n##### Recursive Approach :-\n* Base case: if the root is null than return null\n* Now to connect the left subtree of same level with right subtree of that level\n* The only new line that differentiate from level order traversing ... | 8 | 0 | ['Breadth-First Search', 'Recursion', 'Queue', 'C', 'C++'] | 0 |
populating-next-right-pointers-in-each-node | [JAVA] O(1) Memory Solution (Recursive + Iterative) - Faster than 100% | java-o1-memory-solution-recursive-iterat-myph | 1. Recurisve\n\nclass Solution {\n public Node connect(Node root) {\n if(root == null)\n return root;\n if(root.left != null){\n | Dyanjno123 | NORMAL | 2021-12-29T02:31:31.945808+00:00 | 2022-03-08T02:54:31.309640+00:00 | 769 | false | **1. Recurisve**\n```\nclass Solution {\n public Node connect(Node root) {\n if(root == null)\n return root;\n if(root.left != null){\n if(root.next != null)\n root.right.next = root.next.left;\n else\n root.right.next = null;\n ... | 8 | 2 | ['Java'] | 1 |
populating-next-right-pointers-in-each-node | C++ || O(n) time || O(1) space || No recursion space || 0ms || faster than 100% | c-on-time-o1-space-no-recursion-space-0m-a2yr | The idea is to fix the next pointer for child nodes while traversing through parent node. \nOnce you fixed the next pointer then you can traverse this level in | ajaybedre_07 | NORMAL | 2021-09-16T05:58:58.335848+00:00 | 2021-09-18T18:13:31.974822+00:00 | 150 | false | The idea is to fix the next pointer for child nodes while traversing through parent node. \nOnce you fixed the next pointer then you can traverse this level in O(1) space.\n```\nclass Solution {\npublic:\n Node* connect(Node* root) {\n if(!root)return root;\n Node* curr=root;\n while(curr->left){\n... | 8 | 0 | [] | 0 |
populating-next-right-pointers-in-each-node | python recursion constant space beats 98% short solution | python-recursion-constant-space-beats-98-6h16 | \nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n if not root or not root.left:\n return root\n \n root.le | marriema | NORMAL | 2021-03-18T03:22:30.058767+00:00 | 2021-03-18T03:23:03.443610+00:00 | 451 | false | ```\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n if not root or not root.left:\n return root\n \n root.left.next = root.right\n \n if root.next:\n root.right.next = root.next.left\n \n self.connect(root.left)\n self.... | 8 | 0 | ['Python', 'Python3'] | 3 |
populating-next-right-pointers-in-each-node | 2 different simple recursion O(n) time beats 100% | 2-different-simple-recursion-on-time-bea-ezu5 | The first solution\'s idea is only on about the fact that root.right.next = root.next.left\n\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\' | truongsinh | NORMAL | 2020-03-31T03:17:45.329682+00:00 | 2020-03-31T03:17:45.329733+00:00 | 518 | false | The first solution\'s idea is only on about the fact that `root.right.next = root.next.left`\n```\nclass Solution:\n def connect(self, root: \'Node\') -> \'Node\':\n if not root or not root.left:\n return root\n root.left.next = root.right\n if root.next:\n root.right.next ... | 8 | 0 | ['Recursion', 'Python'] | 0 |
populating-next-right-pointers-in-each-node | 9-line fast c++ without recursion | 9-line-fast-c-without-recursion-by-mitbb-t2n2 | class Solution {\n public:\n void connect(TreeLinkNode *root) {\n while (root) {\n TreeLinkNode *a = root;\n | mitbbs8080 | NORMAL | 2015-11-03T14:02:36+00:00 | 2015-11-03T14:02:36+00:00 | 548 | false | class Solution {\n public:\n void connect(TreeLinkNode *root) {\n while (root) {\n TreeLinkNode *a = root;\n while (a) {\n if (a->left) {\n a->left->next = a->right;\n if (a->next)\n ... | 8 | 2 | [] | 0 |
populating-next-right-pointers-in-each-node | A simple 0 ms recursive solution without helper function | a-simple-0-ms-recursive-solution-without-lhjx | public void connect(TreeLinkNode root) {\n if (root == null){\n return;\n }\n \n if (root.left != null){\n roo | freelikewind | NORMAL | 2015-12-04T02:11:59+00:00 | 2015-12-04T02:11:59+00:00 | 1,329 | false | public void connect(TreeLinkNode root) {\n if (root == null){\n return;\n }\n \n if (root.left != null){\n root.left.next = root.right;\n if (root.next != null){\n root.right.next = root.next.left;\n }\n }\n \n ... | 8 | 1 | ['Java'] | 1 |
populating-next-right-pointers-in-each-node | BFS solution || Explanation || Complexities | bfs-solution-explanation-complexities-by-l3nn | Intuition\nThe goal of this problem is to connect each node at the same level in a perfect binary tree to its immediate neighbor on the right. By establishing t | Anurag_Basuri | NORMAL | 2024-11-02T06:25:26.865113+00:00 | 2024-11-04T13:53:02.457203+00:00 | 1,124 | false | ### Intuition\nThe goal of this problem is to connect each node at the same level in a perfect binary tree to its immediate neighbor on the right. By establishing these connections, each node\u2019s `next` pointer will point to the adjacent node on the same level, or to `None` if it\u2019s the last node in that level. ... | 7 | 0 | ['Linked List', 'Tree', 'Breadth-First Search', 'Binary Tree', 'C++', 'Python3'] | 2 |
populating-next-right-pointers-in-each-node | Python 4 Diffrent Solutions | BFS | DFS | python-4-diffrent-solutions-bfs-dfs-by-p-26hp | 1. Iterative BFS | Space:O(n)\n\n def connect(self, root: \'Optional[Node]\') -> \'Optional[Node]\':\n if( root == None ):\n return root;\n | paramkumar1 | NORMAL | 2023-03-15T05:48:40.114398+00:00 | 2023-03-15T06:56:10.260409+00:00 | 1,865 | false | # 1. Iterative BFS | Space:O(n)\n```\n def connect(self, root: \'Optional[Node]\') -> \'Optional[Node]\':\n if( root == None ):\n return root;\n q = deque();\n q.append(root);\n while( len(q) > 0 ):\n prev = None ;\n size = len(q);\n while(size ... | 7 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Python3'] | 2 |
populating-next-right-pointers-in-each-node | Java | Beauty and Easy solution! | Explained & Visualized | O(n) algo, O(1) memo | | java-beauty-and-easy-solution-explained-6xlwh | The task boils down to understanding the structure of the tree and how to traverse it.\nThe solution is that we create a method that contains the right and left | MykolaMurza | NORMAL | 2022-10-28T17:58:33.720257+00:00 | 2022-10-28T18:02:59.348175+00:00 | 507 | false | The task boils down to understanding the structure of the tree and how to traverse it.\nThe solution is that we create a method that **contains the right and left nodes of the same level**.\n1. We connect the left node with the right one using the `next` field.\n2. The **next** left node of the current left one is comb... | 7 | 0 | ['Depth-First Search', 'Breadth-First Search', 'Recursion', 'Java'] | 1 |
populating-next-right-pointers-in-each-node | Simple Preorder Traversal | simple-preorder-traversal-by-deepak08-z6du | ```class Solution {\n private void helper(Node root)\n {\n if(root == null)\n return;\n //since it is a leaf node it wont have an | deepak08 | NORMAL | 2021-07-24T19:11:54.641206+00:00 | 2021-07-24T19:11:54.641250+00:00 | 226 | false | ```class Solution {\n private void helper(Node root)\n {\n if(root == null)\n return;\n //since it is a leaf node it wont have any child \n if(root.left == null && root.right == null)\n return;\n //stand on parent and connect left child to right child\n roo... | 7 | 0 | ['Java'] | 0 |
populating-next-right-pointers-in-each-node | C Easy Iterative Solution | c-easy-iterative-solution-by-yehudisk-9mo6 | \nstruct Node* connect(struct Node* root) {\n\tif (!root)\n return root;\n \n struct Node* ptr_node = root;\n struct Node* first = ptr_node;\n | yehudisk | NORMAL | 2020-11-13T09:42:31.001653+00:00 | 2020-11-13T09:45:18.671719+00:00 | 414 | false | ```\nstruct Node* connect(struct Node* root) {\n\tif (!root)\n return root;\n \n struct Node* ptr_node = root;\n struct Node* first = ptr_node;\n \n while (ptr_node->left) {\n \n while (ptr_node) {\n \n ptr_node->left->next = ptr_node->right;\n \n ... | 7 | 0 | ['C', 'Iterator'] | 0 |
populating-next-right-pointers-in-each-node | Java recursive solution 100% - O(N) time / O(1) space / O(N) stack call | java-recursive-solution-100-on-time-o1-s-4rbg | Feel free to discuss Big-O analysis :)\njava\nclass Solution {\n public Node connect(Node root) {\n if (root == null) return null;\n if (root.l | monmonpig | NORMAL | 2020-02-14T10:26:50.991781+00:00 | 2020-02-14T10:27:55.787893+00:00 | 311 | false | Feel free to discuss Big-O analysis :)\n```java\nclass Solution {\n public Node connect(Node root) {\n if (root == null) return null;\n if (root.left != null) root.left.next = root.right;\n if (root.right != null) root.right.next = root.next == null ? null : root.next.left;\n connect(root... | 7 | 0 | ['Recursion', 'Java'] | 1 |
populating-next-right-pointers-in-each-node | Python recursive solution | python-recursive-solution-by-he-haitao-h6ii | \u5728\u8FD9\u4E2Arecursion chapter\u5B66\u4E60\u81EA\u5DF1\u609F\u51FA\u6765\u7684\u4E00\u4E2A\u65B9\u6CD5\u3002\n\u4ECEexample\u53EF\u4EE5\u53D1\u73B0\u7ED9\u | he-haitao | NORMAL | 2019-05-13T06:25:00.288923+00:00 | 2019-05-13T07:25:49.548166+00:00 | 715 | false | \u5728\u8FD9\u4E2Arecursion chapter\u5B66\u4E60\u81EA\u5DF1\u609F\u51FA\u6765\u7684\u4E00\u4E2A\u65B9\u6CD5\u3002\n\u4ECEexample\u53EF\u4EE5\u53D1\u73B0\u7ED9\u7684\u89C4\u5F8B\u5C31\u662F\n\u6BCF\u4E2A\u8282\u70B9\uFF08\u975E\u53F6\u5B50\u7ED3\u70B9\uFF09\u5FC5\u6709\u4E24\u4E2A\u5B50\u8282\u70B9\n\u5BF9\u6BCF\u4E00le... | 7 | 0 | [] | 1 |
populating-next-right-pointers-in-each-node | C# while | c-while-by-bacon-f68w | \npublic class Solution {\n public Node Connect(Node root) {\n var preMostLeft = root;\n while (preMostLeft != null) {\n var cur = p | bacon | NORMAL | 2019-05-05T00:36:19.258288+00:00 | 2019-05-05T00:36:19.258369+00:00 | 584 | false | ```\npublic class Solution {\n public Node Connect(Node root) {\n var preMostLeft = root;\n while (preMostLeft != null) {\n var cur = preMostLeft;\n while (cur != null && cur.left != null) {\n cur.left.next = cur.right;\n if (cur.next != null) {\n ... | 7 | 0 | [] | 0 |
populating-next-right-pointers-in-each-node | Java/c++/c solutions. Elegant iterative solutions (4 lines) and recursive solutions (5 lines) | javacc-solutions-elegant-iterative-solut-v3gg | C++\n## Recursive\nSetting the next pointer from the parent node makes for a clean solution. The only thing to worry about is how to handle the far right node o | christrompf | NORMAL | 2018-08-25T09:03:33.073515+00:00 | 2022-05-13T05:01:10.025914+00:00 | 457 | false | # **C++**\n## Recursive\nSetting the _next_ pointer from the parent node makes for a clean solution. The only thing to worry about is how to handle the far right node of each level, which doesn\'t have a _next_. This is easy to check for too, just check if the parent node has a _next_ pointer, if it doesn\'t, then you\... | 7 | 0 | ['C', 'Java'] | 1 |
populating-next-right-pointers-in-each-node | My recursive solution | my-recursive-solution-by-jwang8-8ij2 | void connect(TreeLinkNode *root) {\n if( root == NULL || root->left == NULL && root->right == NULL ) //{} \\ {0}\n {\n return;\n | jwang8 | NORMAL | 2014-07-14T17:30:25+00:00 | 2014-07-14T17:30:25+00:00 | 2,190 | false | void connect(TreeLinkNode *root) {\n if( root == NULL || root->left == NULL && root->right == NULL ) //{} \\ {0}\n {\n return;\n }\n \n TreeLinkNode *p, *q;\n p = root->left;\n q = root->right;\n p->next = q;\n while( p->right != NULL )... | 7 | 0 | [] | 6 |
populating-next-right-pointers-in-each-node | Simple Iterative solution | simple-iterative-solution-by-1337beef-xnc4 | Populate the levels one by one. curLevel points to node whose children will be linked, nextLevel points to the first node in the next level.\n\n class Soluti | 1337beef | NORMAL | 2014-09-22T15:46:19+00:00 | 2014-09-22T15:46:19+00:00 | 1,154 | false | Populate the levels one by one. curLevel points to node whose children will be linked, nextLevel points to the first node in the next level.\n\n class Solution {\n public:\n void connect(TreeLinkNode *root) {\n if(!root)return;\n TreeLinkNode*curLevel=root,*nextLevel=root->left;\n ... | 7 | 0 | [] | 0 |
populating-next-right-pointers-in-each-node | Sharing my Java O(1) extra space code | sharing-my-java-o1-extra-space-code-by-w-yrv6 | public void connect(TreeLinkNode root) {\n TreeLinkNode cur;\n TreeLinkNode nextLevel = root;\n while (nextLevel != null) {\n cu | wzhang84 | NORMAL | 2015-04-05T14:22:09+00:00 | 2015-04-05T14:22:09+00:00 | 952 | false | public void connect(TreeLinkNode root) {\n TreeLinkNode cur;\n TreeLinkNode nextLevel = root;\n while (nextLevel != null) {\n cur = nextLevel;\n // at each level, connects the children nodes\n while (cur != null && \n cur.left != null // check... | 7 | 0 | ['Java'] | 0 |
populating-next-right-pointers-in-each-node | Another accepted Java solution | another-accepted-java-solution-by-jeanti-ub16 | Basically, we use the next pointer to help level traversal. No recursion is needed, O(1) constant space, O(n) running time.\n\n public class Solution {\n | jeantimex | NORMAL | 2015-07-04T00:50:28+00:00 | 2015-07-04T00:50:28+00:00 | 644 | false | Basically, we use the next pointer to help level traversal. No recursion is needed, O(1) constant space, O(n) running time.\n\n public class Solution {\n public void connect(TreeLinkNode root) {\n if (root == null) return;\n \n while (root.left != null) {\n Tree... | 7 | 0 | ['Java'] | 0 |
populating-next-right-pointers-in-each-node | Beats 96.94% with step by step explanation, | beats-9694-with-step-by-step-explanation-ar7v | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nThis solution uses a level-order traversal of the binary tree and a queue | Marlen09 | NORMAL | 2023-02-17T04:20:09.620749+00:00 | 2023-02-17T04:20:09.620790+00:00 | 1,486 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThis solution uses a level-order traversal of the binary tree and a queue to keep track of the nodes. For each level, it sets the next pointers of the nodes to the next node in the queue, except for the last node in the leve... | 6 | 0 | ['Tree', 'Depth-First Search', 'Breadth-First Search', 'Python', 'Python3'] | 3 |
populating-next-right-pointers-in-each-node | ✅ [Python, C++,Java]|| Beginner level Solution||100%Faster||Simple-Short-Solution✅ | python-cjava-beginner-level-solution100f-k4su | Please upvote to motivate me in my quest of documenting all leetcode solutions. HAPPY CODING:)\nAny suggestions and improvements are always welcome.\n___\n__\nQ | Anos | NORMAL | 2022-08-14T10:36:01.465570+00:00 | 2022-08-14T10:36:01.465604+00:00 | 442 | false | ***Please upvote to motivate me in my quest of documenting all leetcode solutions. HAPPY CODING:)\nAny suggestions and improvements are always welcome*.**\n___________________\n_________________\n***Q116. Populating Next Right Pointers in Each Node***\n\nYou are given a perfect binary tree where all leaves are on the s... | 6 | 0 | ['Breadth-First Search', 'C', 'Python', 'Java'] | 0 |
populating-next-right-pointers-in-each-node | C++ || 5 lines code || Very Easy || DFS || Simple || Explanation | c-5-lines-code-very-easy-dfs-simple-expl-z3rc | This solution implements a recursive function - it takes a node, connects the NEXT pointers of it\'s left & right children, and repeats the process for the chil | manoharbanda_ | NORMAL | 2022-03-01T06:56:39.833677+00:00 | 2022-03-01T07:02:29.692019+00:00 | 258 | false | This solution implements a recursive function - it takes a node, connects the NEXT pointers of it\'s left & right children, and repeats the process for the children. So the algorithm is :\nFor each non-leaf node, we do -\n(1) NEXT of Left child is pointed to Right child. Easy.\n(2) NEXT of Right child is pointed to NEX... | 6 | 0 | ['Depth-First Search', 'Recursion', 'C'] | 1 |
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