post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2338726/Python3-recursive-approach | class Solution:
def __init__(self):
self.prev = None
def flatten(self, root):
if not root:
return None
self.flatten(root.right)
self.flatten(root.left)
root.right = self.prev
root.left = None
self.prev = root | flatten-binary-tree-to-linked-list | π Python3 recursive approach | Dark_wolf_jss | 6 | 360 | flatten binary tree to linked list | 114 | 0.612 | Medium | 900 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/1608790/Pythonor-Simple-and-Easy-2-approachesor-Faster-than-98.13 | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
#writing the function to flatten the binary tree
def helper(root):
"""
param type root: TreeNode
... | flatten-binary-tree-to-linked-list | Python| Simple and Easy 2 approaches| Faster than 98.13% | lunarcrab | 3 | 181 | flatten binary tree to linked list | 114 | 0.612 | Medium | 901 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/1608790/Pythonor-Simple-and-Easy-2-approachesor-Faster-than-98.13 | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
res=[]
# Functtion to return the preorder Traversal
def helper(root):
if not root:
return
... | flatten-binary-tree-to-linked-list | Python| Simple and Easy 2 approaches| Faster than 98.13% | lunarcrab | 3 | 181 | flatten binary tree to linked list | 114 | 0.612 | Medium | 902 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/750312/Python3-Solution-with-a-Detailed-Explanation-Flatten-binary-tree-to-linkedlist | class Solution:
def flatten(self, root):
if not root: #1
return
if root.left: #2
self.flatten(root.left) #3
temp = root.right #4
root.right = root.left #5
root.left = None #6
curr = root.right #7
... | flatten-binary-tree-to-linked-list | Python3 Solution with a Detailed Explanation - Flatten binary tree to linkedlist | peyman_np | 2 | 205 | flatten binary tree to linked list | 114 | 0.612 | Medium | 903 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2345900/Py-or-2-iterative-methods-or-O(1)-space | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
if not root: return
prev = None
stack = [root]
while stack:
node = stack.pop()
if node.right: stack.appe... | flatten-binary-tree-to-linked-list | Py | 2 iterative methods | O(1) space | XUEYANZ | 1 | 63 | flatten binary tree to linked list | 114 | 0.612 | Medium | 904 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2040103/O(n)timeBEATS-99.97-MEMORYSPEED-0ms-MAY-2022 | class Solution:
def flatten(self, root: TreeNode) -> None:
head, curr = None, root
while head != root:
if curr.right == head: curr.right = None
if curr.left == head: curr.left = None
if curr.right: curr = curr.right
elif curr.left: curr = curr.left
... | flatten-binary-tree-to-linked-list | O(n)time/BEATS 99.97% MEMORY/SPEED 0ms MAY 2022 | cucerdariancatalin | 1 | 98 | flatten binary tree to linked list | 114 | 0.612 | Medium | 905 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/1603835/Python-3-line-recursive | class Solution:
def flatten(self, root: Optional[TreeNode], right=None) -> None:
if not root: return right
root.left, root.right = None, self.flatten(root.left, self.flatten(root.right, right))
return root | flatten-binary-tree-to-linked-list | [Python] 3-line recursive | hooneken | 1 | 117 | flatten binary tree to linked list | 114 | 0.612 | Medium | 906 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/1603835/Python-3-line-recursive | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
cur = root
while cur:
left = cur.left
if left:
while left.right: left = left.right
left.right = cur.right
cur.left, cur.right = None, cur.left
c... | flatten-binary-tree-to-linked-list | [Python] 3-line recursive | hooneken | 1 | 117 | flatten binary tree to linked list | 114 | 0.612 | Medium | 907 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/1474276/Python-iterative-solution-using-preorder | class Solution:
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
# curr = TreeNode(0)
if not root:
return None
stack = [root]
while len(stack)>0:
temp = stack.pop()
# here, firs... | flatten-binary-tree-to-linked-list | Python iterative solution, using preorder | Ridzzz | 1 | 170 | flatten binary tree to linked list | 114 | 0.612 | Medium | 908 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/1398415/Python-Clean-and-Easy-or-92.29-speed | class Solution:
def flatten(self, root: TreeNode) -> None:
if not root: return
queue = deque([root.right, root.left])
while queue:
node = queue.pop()
if not node: continue
root.left, root.right = None, node
root = root.right
queue.a... | flatten-binary-tree-to-linked-list | [Python] Clean and Easy | 92.29% speed | soma28 | 1 | 155 | flatten binary tree to linked list | 114 | 0.612 | Medium | 909 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/696054/Python-DFS-Solution-Easy-to-Understand | class Solution:
# Time: O(n)
# Space: O(H)
def flatten(self, node: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
if not node:
return
self.flatten(node.left)
self.flatten(node.right)
if not node.right and n... | flatten-binary-tree-to-linked-list | Python DFS Solution Easy to Understand | whissely | 1 | 156 | flatten binary tree to linked list | 114 | 0.612 | Medium | 910 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/692703/Python3-5-line-recursive | class Solution:
def flatten(self, root: TreeNode) -> None:
def fn(node, tail=None):
"""Return head of flattened binary tree"""
if not node: return tail
node.left, node.right = None, fn(node.left, fn(node.right, tail))
return node
fn... | flatten-binary-tree-to-linked-list | [Python3] 5-line recursive | ye15 | 1 | 109 | flatten binary tree to linked list | 114 | 0.612 | Medium | 911 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2730476/Striver-approach | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
prev = None
def solve(node):
nonlocal prev
if not node:
return None
solve(node.right)
solve(node.left)
node.right = prev
node.left = ... | flatten-binary-tree-to-linked-list | Striver approach | hacktheirlives | 0 | 4 | flatten binary tree to linked list | 114 | 0.612 | Medium | 912 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2688353/python-or-recursion | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
def helper(root):
if not root:
return None
# get leaf
left = helper(root.left)
right = helper(root.right)
# left leaf move to right subt... | flatten-binary-tree-to-linked-list | python | recursion | MichelleZou | 0 | 9 | flatten binary tree to linked list | 114 | 0.612 | Medium | 913 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2459663/Python-Recursive-Approach | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
def dfs(root):
while root:
if root.left:
dfs(root.left)
node1 = root.left
while node1.right:
node1 = node1.right
... | flatten-binary-tree-to-linked-list | Python Recursive Approach | Abhi_009 | 0 | 35 | flatten binary tree to linked list | 114 | 0.612 | Medium | 914 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2346551/Easy-peasy-Solution-in-Python-3-which-cannot-be-difficult | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
if not root: return root
elif not root.left and not root.right: return root
stack, nums = [root], []
while stack:
node = stack.pop()
if node:
nums.append(node)
stack.append(node.right)
stack.append(node.left)
cursor ... | flatten-binary-tree-to-linked-list | Easy-peasy Solution in Python 3, which cannot be difficult | HappyLunchJazz | 0 | 38 | flatten binary tree to linked list | 114 | 0.612 | Medium | 915 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2342956/Python-Bruteforce-With-Preorder-Traversal-Helper | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
preorder = []
self.helper(root, preorder)
for node in preorder[1:]:
root.left = None
root.right = node
root = root.right
def helper(self, root,... | flatten-binary-tree-to-linked-list | Python Bruteforce With Preorder Traversal Helper | HunkWhoCodes | 0 | 7 | flatten binary tree to linked list | 114 | 0.612 | Medium | 916 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2342478/Python-solution-or-Flatten-Binary-Tree-to-Linked-List | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
if not root:
return None
left = self.flatten(root.left)
right = self.flatten(root.right)
if root.left:
left.right = root.right
root.right = root.left
root.left = None
... | flatten-binary-tree-to-linked-list | Python solution | Flatten Binary Tree to Linked List | nishanrahman1994 | 0 | 15 | flatten binary tree to linked list | 114 | 0.612 | Medium | 917 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2342382/Easy-Python-With-Explanation-DFS-wRecursion | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
# Depth-first recursion
def dfs(node):
# Recursion base case
# do nothing if none
if not node:
... | flatten-binary-tree-to-linked-list | Easy Python With Explanation DFS w/Recursion | drblessing | 0 | 17 | flatten binary tree to linked list | 114 | 0.612 | Medium | 918 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2342018/Python-Simple-Python-Solution-Using-Depth-First-Search | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
self.previous_node = None
def DepthFirstSearch(root):
if root is None:
return None
DepthFirstSearch(root.right)
DepthFirstSearch(root.left)
root.right = sel... | flatten-binary-tree-to-linked-list | [ Python ] β
β
Simple Python Solution Using Depth First Search π₯β π₯³π | ASHOK_KUMAR_MEGHVANSHI | 0 | 27 | flatten binary tree to linked list | 114 | 0.612 | Medium | 919 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2341111/Very-simple-and-short-python-code | class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
def flat... | flatten-binary-tree-to-linked-list | Very simple and short python code | pradyumna04 | 0 | 5 | flatten binary tree to linked list | 114 | 0.612 | Medium | 920 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2341053/Beginner's-Solution-Python | class linkedlist:
def __init__(self):
self.head = None
self.tail = None
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
if root == None:
return root
result = self.helper(root)
return result.head
... | flatten-binary-tree-to-linked-list | Beginner's Solution - Python | niteshjyo | 0 | 18 | flatten binary tree to linked list | 114 | 0.612 | Medium | 921 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2340023/Python-39ms-oror-Simple-Recursive-2-Tail-Pointers-Space-O(1)-oror-Documented | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
# flatten given root of binary tree and return the tail of linked list
def recursive(node) -> Optional[TreeNode]:
if node:
# these will be leaf nodes of the given subtrees/linked list
tai... | flatten-binary-tree-to-linked-list | [Python] 39ms || Simple Recursive 2-Tail Pointers - Space O(1) || Documented | Buntynara | 0 | 4 | flatten binary tree to linked list | 114 | 0.612 | Medium | 922 |
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/discuss/2339175/Python-simple-commented-recursive-solution | class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
if not root:
return None
# left is the last node of the left subtree.
# and similarly right is the last node of the ... | flatten-binary-tree-to-linked-list | Python simple commented recursive solution | kaustav43 | 0 | 12 | flatten binary tree to linked list | 114 | 0.612 | Medium | 923 |
https://leetcode.com/problems/distinct-subsequences/discuss/1472969/Python-Bottom-up-DP-Explained | class Solution:
def numDistinct(self, s: str, t: str) -> int:
m = len(s)
n = len(t)
dp = [[0] * (n+1) for _ in range(m+1)]
for i in range(m+1):
dp[i][0] = 1
"""redundant, as we have initialised dp table with full of zeros"""
# for i in ra... | distinct-subsequences | Python - Bottom up DP - Explained | ajith6198 | 6 | 475 | distinct subsequences | 115 | 0.439 | Hard | 924 |
https://leetcode.com/problems/distinct-subsequences/discuss/2040129/O(n)timeBEATS-99.97-MEMORYSPEED-0ms-MAY-2022 | class Solution:
def numDistinct(self, s: str, t: str) -> int:
## RC ##
## APPROACH : DP ##
## LOGIC ##
# 1. Point to be noted, empty seq is also subsequence of any sequence i.e "", "" should return 1. so we fill the first row accordingly
# 2. if chars match, we get the maximum of [diagonal + upper, ... | distinct-subsequences | O(n)time/BEATS 99.97% MEMORY/SPEED 0ms MAY 2022 | cucerdariancatalin | 4 | 231 | distinct subsequences | 115 | 0.439 | Hard | 925 |
https://leetcode.com/problems/distinct-subsequences/discuss/2775154/Linear-order-Python-Easy-Solution | class Solution:
def numDistinct(self, s: str, t: str) -> int:
def disSub(s,t,ind_s,ind_t,dp):
if ind_t<0:
return 1
if ind_s<0:
return 0
if s[ind_s]==t[ind_t]:
if (dp[ind_t])[ind_s]!=-1:
return (dp[ind_t... | distinct-subsequences | Linear order Python Easy Solution | Khan_Baba | 2 | 176 | distinct subsequences | 115 | 0.439 | Hard | 926 |
https://leetcode.com/problems/distinct-subsequences/discuss/2775154/Linear-order-Python-Easy-Solution | class Solution:
def numDistinct(self, s: str, t: str) -> int:
ls=len(s)
lt=len(t)
dp=[[0]*(ls+1) for i in range(lt+1)]
for i in range(ls):
(dp[0])[i]=1
for ind_t in range(1,lt+1):
for ind_s in range (1,ls+1):
if s[ind_s-1]==t[ind_t-1]:... | distinct-subsequences | Linear order Python Easy Solution | Khan_Baba | 2 | 176 | distinct subsequences | 115 | 0.439 | Hard | 927 |
https://leetcode.com/problems/distinct-subsequences/discuss/2775154/Linear-order-Python-Easy-Solution | class Solution:
def numDistinct(self, s: str, t: str) -> int:
ls=len(s)
lt=len(t)
prev=[1]*(ls+1)
for ind_t in range(1,lt+1):
temp=[0]*(ls+1)
for ind_s in range (1,ls+1):
if s[ind_s-1]==t[ind_t-1]:
... | distinct-subsequences | Linear order Python Easy Solution | Khan_Baba | 2 | 176 | distinct subsequences | 115 | 0.439 | Hard | 928 |
https://leetcode.com/problems/distinct-subsequences/discuss/2437315/Python-DP-O(m*n)-Time | class Solution:
def numDistinct(self, s: str, t: str) -> int:
dp = [[0] * (len(s)+1) for _ in range(len(t)+1)]
# 0th row => len(t) == 0 and len(s) can be anything. So empty string t can be Subsequence of s
for j in range(len(s)+1):
dp[0][j] = 1
for i in ... | distinct-subsequences | Python - DP - O(m*n) Time | samirpaul1 | 1 | 44 | distinct subsequences | 115 | 0.439 | Hard | 929 |
https://leetcode.com/problems/distinct-subsequences/discuss/580112/DP-Solution-(Beats-99-in-Time-58-in-Space) | class Solution:
def numDistinct(self, s: str, t: str) -> int:
c_indices = collections.defaultdict(list)
max_counts = [1] + [0] * len(t)
for i, c in enumerate(t):
c_indices[c].append(i + 1)
for c in s:
for i in reversed(c_indices[c]):
max_count... | distinct-subsequences | DP Solution (Beats 99% in Time, 58% in Space) | skarakulak | 1 | 208 | distinct subsequences | 115 | 0.439 | Hard | 930 |
https://leetcode.com/problems/distinct-subsequences/discuss/2805876/Python-bottom-up-approach | class Solution:
def numDistinct(self, s: str, t: str) -> int:
n = len(s)
m = len(t)
if n<m:
return 0
dp = [1]+[0]*m
for i in range(n):
for j in reversed(range(m)):
if s[i]==t[j]:
dp[j+1]+=dp[j]
return dp[m] | distinct-subsequences | Python bottom up approach | Akshit-Gupta | 0 | 2 | distinct subsequences | 115 | 0.439 | Hard | 931 |
https://leetcode.com/problems/distinct-subsequences/discuss/2784680/Python(Tabulation) | class Solution:
def numDistinct(self, s: str, t: str) -> int:
dp=[[0]*(len(t)+1) for i in range(len(s)+1)]
for i in range(len(s)+1):
dp[i][0]=1
for i in range(1,len(s)+1):
for j in range(1,len(t)+1):
if s[i-1]==t[j-1]:
dp[i][j] = dp... | distinct-subsequences | Python(Tabulation) | parasgarg31 | 0 | 4 | distinct subsequences | 115 | 0.439 | Hard | 932 |
https://leetcode.com/problems/distinct-subsequences/discuss/2784674/Simple-Python-Solution(Striver) | class Solution:
def numDistinct(self, s: str, t: str) -> int:
dp=[[-1]*len(t) for i in range(len(s))]
def solve(ind1,ind2):
if ind2<0:
return 1
if ind1<0:
return 0
if dp[ind1][ind2]!=-1:
return dp[ind1][... | distinct-subsequences | Simple Python Solution(Striver) | parasgarg31 | 0 | 3 | distinct subsequences | 115 | 0.439 | Hard | 933 |
https://leetcode.com/problems/distinct-subsequences/discuss/2760304/Python-Three-approach-recursion-top-down-bottom-up | class Solution:
def solve(self , s , t, i , j):
if j == 0:
return 1
if i == 0:
return 0
if s[i-1] == t[j-1]:
take = self.solve(s , t, i-1 , j-1)
move = self.solve(s , t , i-1 , j)
return (take + move)
else:
ret... | distinct-subsequences | Python Three approach recursion , top-down , bottom-up | rajitkumarchauhan99 | 0 | 4 | distinct subsequences | 115 | 0.439 | Hard | 934 |
https://leetcode.com/problems/distinct-subsequences/discuss/2694849/PYTHONoror-EASY-SOLUoror-with-EXPLAINATION-using-DP-Bottomup-approach | class Solution:
def numDistinct(self, s: str, t: str) -> int:
m=len(s)
n=len(t)
dp=[]
for i in range (m+1): #making dp of m+1 and n+1
dp.append([0]*(n+1))
dp[0][0]=1 #making 1st element of 2d dp as 1
for i in range (m+1): #making 1st column of ... | distinct-subsequences | PYTHON|| EASY SOLU|| with EXPLAINATION using DP Bottomup approach | tush18 | 0 | 8 | distinct subsequences | 115 | 0.439 | Hard | 935 |
https://leetcode.com/problems/distinct-subsequences/discuss/2431193/Clean-Fast-Python3-or-Bottom-Up-DP-or-O(s-%2B-t) | class Solution:
def numDistinct(self, s: str, t: str) -> int:
ls, lt = len(s), len(t)
if ls < lt:
return 0
# dp
subseqs = [[0] * (ls + 1) for _ in range(lt + 1)]
subseqs[0] = [1] * (ls + 1)
letters_seen = set()
for ti in range(lt):
... | distinct-subsequences | Clean, Fast Python3 | Bottom Up DP | O(s + t) | ryangrayson | 0 | 9 | distinct subsequences | 115 | 0.439 | Hard | 936 |
https://leetcode.com/problems/distinct-subsequences/discuss/2429305/Memoization-and-Recursion | class Solution:
def numDistinct(self, s: str, t: str) -> int:
n, m = len(s), len(t)
lookup = {}
def f(i, j, lookup):
if j<0: return 1
if i<0: return 0
if (i, j) in lookup: return lookup[(i,j)]
#matchi... | distinct-subsequences | Memoization & Recursion | logeshsrinivasans | 0 | 16 | distinct subsequences | 115 | 0.439 | Hard | 937 |
https://leetcode.com/problems/distinct-subsequences/discuss/2429305/Memoization-and-Recursion | class Solution:
def numDistinct(self, s: str, t: str) -> int:
m, n = len(s), len(t)
dp = [[0] * (n+1) for i in range(m+1)]
for i in range(m+1):
dp[i][0] = 1
for i in range(1, m+1):
for j in range(1, n+1):
if s[i-1] == t[j-... | distinct-subsequences | Memoization & Recursion | logeshsrinivasans | 0 | 16 | distinct subsequences | 115 | 0.439 | Hard | 938 |
https://leetcode.com/problems/distinct-subsequences/discuss/2420912/Python-DP-Solution | class Solution:
def numDistinct(self, s: str, t: str) -> int:
d = {}
def r(index_s, index_t):
if index_t < 0:
return 1
if index_s < 0:
return 0
if s[index_s] == t[index_t]:
if (index_s-1, index_t-1) not in d:
... | distinct-subsequences | Python DP Solution | DietCoke777 | 0 | 17 | distinct subsequences | 115 | 0.439 | Hard | 939 |
https://leetcode.com/problems/distinct-subsequences/discuss/2341291/Python-Solution-or-Recursion-or-DP | class Solution:
def numDistinct(self, s: str, t: str) -> int:
#Memoiation
n=len(s)
m=len(t)
dp=[[-1]*(m+1) for i in range(n+1)]
def helper(i, j):
if j<0:
return 1
if i<0:
return 0
if dp[i][j]!=-1:
... | distinct-subsequences | Python Solution | Recursion | DP | Siddharth_singh | 0 | 40 | distinct subsequences | 115 | 0.439 | Hard | 940 |
https://leetcode.com/problems/distinct-subsequences/discuss/2335067/Python-easy-to-read-and-understand-or-dynamic-programming | class Solution:
def solve(self, x, y, m, n):
if n == 0:
return 1
elif m == 0:
return 0
if x[m-1] == y[n-1]:
return self.solve(x, y, m-1, n-1) + self.solve(x, y, m-1, n)
else:
return self.solve(x, y, m-1, n)
def numDistinct(self... | distinct-subsequences | Python easy to read and understand | dynamic-programming | sanial2001 | 0 | 21 | distinct subsequences | 115 | 0.439 | Hard | 941 |
https://leetcode.com/problems/distinct-subsequences/discuss/2335067/Python-easy-to-read-and-understand-or-dynamic-programming | class Solution:
def solve(self, x, y, m, n):
if n == 0:
return 1
elif m == 0:
return 0
if (m, n) in self.d:
return self.d[(m, n)]
if x[m-1] == y[n-1]:
self.d[(m, n)] = self.solve(x, y, m-1, n-1) + self.solve(x, y, m-1, n)
re... | distinct-subsequences | Python easy to read and understand | dynamic-programming | sanial2001 | 0 | 21 | distinct subsequences | 115 | 0.439 | Hard | 942 |
https://leetcode.com/problems/distinct-subsequences/discuss/2301931/Python3-oror-My-first-DP-thought-solution-ever-83.86-of-Python3-submissions | class Solution:
def numDistinct(self, s: str, t: str) -> int:
r, c = len(t), len(s)
dp = [[0]*c for _ in range(r)]
for i in range(c):
dp[0][i] = 1 + dp[0][i-1] if t[0] == s[i] else dp[0][i-1]
for i in range(1, r):
for j in range(1, c):
dp[i][j] = dp[i-1][j-1] + dp[i][j-1] if t[i] == s[j] else dp[i... | distinct-subsequences | Python3 || My first DP thought solution ever 83.86% of Python3 submissions | sagarhasan273 | 0 | 43 | distinct subsequences | 115 | 0.439 | Hard | 943 |
https://leetcode.com/problems/distinct-subsequences/discuss/2244692/easy-to-understand-Memoized-and-Bottom-up-solution | class Solution:
def numDistinct(self, s: str, t: str) -> int:
# Bottom-up solution
# T(c)=O(n*m)
# S(c)=O(n*m)
n=len(s)
m=len(t)
dp=[[0 for _ in range(m+1)] for _ in range(n+1)]
for i in range(n+1):
for j in range(m+1):
... | distinct-subsequences | easy to understand Memoized and Bottom-up solution | Aniket_liar07 | 0 | 29 | distinct subsequences | 115 | 0.439 | Hard | 944 |
https://leetcode.com/problems/distinct-subsequences/discuss/2228316/DP-oror-Top-Down-oror-Memoization | class Solution:
def distinctSubsequences(self, i, j, s, t, dpCache):
if j == len(t):
return 1
if i == len(s):
return 0
if (i, j) in dpCache:
return dpCache[(i, j)]
if s[i] == t[j]:
dpCache[(i, j)] = self.distinctSubsequences(i ... | distinct-subsequences | DP || Top Down || Memoization | Vaibhav7860 | 0 | 36 | distinct subsequences | 115 | 0.439 | Hard | 945 |
https://leetcode.com/problems/distinct-subsequences/discuss/2190935/Python3-clean-code-(recursion-%2B-memoization) | class Solution:
def numDistinct(self, s: str, t: str) -> int:
def get_subsequences(s, t, index_1, index_2, memo):
if index_1 == len(s):
if index_2 == len(t):
return 1
return 0
if index_2 == len(t):
return 1
... | distinct-subsequences | Python3 clean code (recursion + memoization) | myvanillaexistence | 0 | 35 | distinct subsequences | 115 | 0.439 | Hard | 946 |
https://leetcode.com/problems/distinct-subsequences/discuss/1686168/step-by-step-explanation-of-dp-solution-from-O(mn)-space-to-O(n)-space-O(mn)time | class Solution:
def numDistinct(self, s: str, t: str) -> int:
m,n=len(s),len(t)
dp=[[0 for _ in range(n+1)] for _ in range(m+1)]
for i in range(m+1):
dp[i][n] = 1
dp[-1][-1]=1
for i in range(m-1,-1,-1):
for j in range(n-1,-1,-1):
dp[i][j] = dp[i+1][j] # not using s[i]
if s[i]==t[j]:
... | distinct-subsequences | step-by-step explanation of dp solution from O(mn) space to O(n) space, O(mn)time | jacot | 0 | 70 | distinct subsequences | 115 | 0.439 | Hard | 947 |
https://leetcode.com/problems/distinct-subsequences/discuss/1686168/step-by-step-explanation-of-dp-solution-from-O(mn)-space-to-O(n)-space-O(mn)time | class Solution:
def numDistinct(self, s: str, t: str) -> int:
m,n=len(s),len(t)
dp=[[0 for _ in range(n+1)] for _ in range(2)]
for i in range(2):
dp[i][n] = 1
for i in range(m-1,-1,-1):
for j in range(n-1,-1,-1):
dp[0][j] = dp[1][j] # not using s[i]
if s[i]==t[j]:
dp[0][j] += dp[1][j+1] # u... | distinct-subsequences | step-by-step explanation of dp solution from O(mn) space to O(n) space, O(mn)time | jacot | 0 | 70 | distinct subsequences | 115 | 0.439 | Hard | 948 |
https://leetcode.com/problems/distinct-subsequences/discuss/1653663/Python3-Easy-DFS-%2B-Memoization | class Solution:
def numDistinct(self, s: str, t: str) -> int:
n, m = len(s), len(t)
@cache
def dfs(i, j):
if (j >= m):
return 1
if (i >= n):
return 0
if (s[i] == t[j]):
return dfs(i + 1,... | distinct-subsequences | Python3 - Easy DFS + Memoization β
| Bruception | 0 | 175 | distinct subsequences | 115 | 0.439 | Hard | 949 |
https://leetcode.com/problems/distinct-subsequences/discuss/1472826/Distinct-Subsequences-oror-DP | class Solution:
def numDistinct(self, s: str, t: str) -> int:
memo = {}
def dfs(i, j):
if (i,j) in memo:
return memo[(i,j)]
if j == len(t):
return 1
if i == len(s) or len(t)-j > len(s)-i:
return 0
... | distinct-subsequences | Distinct Subsequences || DP | mdAzhar | 0 | 59 | distinct subsequences | 115 | 0.439 | Hard | 950 |
https://leetcode.com/problems/distinct-subsequences/discuss/1309048/2-Pointer-Recursion-2D-1D-iteration-Edit-distance-delete-operation-) | class Solution:
def numDistinctTopDown(self, s: str, t: str) -> int:
# this is like edit distance... with only delete ;)
@functools.cache
def ways(si, ti): # index in both strings
if ti == len(t):
return 1
elif si == len(s):
return 0
... | distinct-subsequences | 2 Pointer, Recursion, 2D, 1D iteration Edit distance delete operation ;) | yozaam | 0 | 76 | distinct subsequences | 115 | 0.439 | Hard | 951 |
https://leetcode.com/problems/distinct-subsequences/discuss/692969/Python3-dynamic-programming | class Solution:
def numDistinct(self, s: str, t: str) -> int:
@lru_cache(None)
def fn(i, j):
"""Return number of subsequences of s[i:] which equals t[j:]"""
if j == len(t): return 1
if i == len(s): return 0
return fn(i+1, j) + (s[i] == t[j]) ... | distinct-subsequences | [Python3] dynamic programming | ye15 | 0 | 52 | distinct subsequences | 115 | 0.439 | Hard | 952 |
https://leetcode.com/problems/distinct-subsequences/discuss/692969/Python3-dynamic-programming | class Solution:
def numDistinct(self, s: str, t: str) -> int:
loc = {}
for i, x in enumerate(t): loc.setdefault(x, []).append(i)
ans = [0]*len(t) + [1]
for c in reversed(s):
for i in pos.get(c, []): ans[i] += ans[i+1]
return ans[0] | distinct-subsequences | [Python3] dynamic programming | ye15 | 0 | 52 | distinct subsequences | 115 | 0.439 | Hard | 953 |
https://leetcode.com/problems/distinct-subsequences/discuss/445405/python-solution-recursive-with-memory-beats-99 | class Solution:
def numDistinct(self, s: str, t: str) -> int:
nextp = [ {} for _ in range(len(s))]
current={}
for i in range(len(s)-1,-1,-1):
current[s[i]] = i
nextp[i]=current.copy()
self.mem={}
return self.helper(s,0,t,0,nextp)
def helper(self,s... | distinct-subsequences | python solution recursive with memory, beats 99% | meowthecat | 0 | 88 | distinct subsequences | 115 | 0.439 | Hard | 954 |
https://leetcode.com/problems/distinct-subsequences/discuss/1226056/Python-O(n2)-solution-using-DP | class Solution:
def numDistinct(self, s: str, t: str) -> int:
m, n = len(s), len(t)
dp = [[0 for i in range(m + 1)] for i in range(n + 1)]
for i in range(0, m + 1):
dp[i][0] = 1
for i in range(1,m + 1):
for j in range(1,n + 1):
dp[i][j... | distinct-subsequences | Python O(n^2) solution using DP | m0biu5 | -1 | 95 | distinct subsequences | 115 | 0.439 | Hard | 955 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/719347/Python-Solution-O(1)-and-O(n)-memory. | class Solution:
def connect(self, root: 'Node') -> 'Node':
# edge case check
if not root:
return None
# initialize the queue with root node (for level order traversal)
queue = collections.deque([root])
# start the traversal
while queue:
... | populating-next-right-pointers-in-each-node | Python Solution O(1) and O(n) memory. | darshan_22 | 10 | 534 | populating next right pointers in each node | 116 | 0.596 | Medium | 956 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/719347/Python-Solution-O(1)-and-O(n)-memory. | class Solution:
def connect(self, root: 'Node') -> 'Node':
# edge case check
if not root:
return None
node = root # create a pointer to the root node
# iterate only until we have a new level (because the connections for Nth level are done when we are at ... | populating-next-right-pointers-in-each-node | Python Solution O(1) and O(n) memory. | darshan_22 | 10 | 534 | populating next right pointers in each node | 116 | 0.596 | Medium | 957 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2040150/O(n)timeBEATS-99.97-MEMORYSPEED-0ms-MAY-2022 | class Solution:
def connect(self, root: 'Node') -> 'Node':
# edge case check
if not root:
return None
node = root # create a pointer to the root node
# iterate only until we have a new level (because the connections for Nth level are done when we are at ... | populating-next-right-pointers-in-each-node | O(n)time/BEATS 99.97% MEMORY/SPEED 0ms MAY 2022 | cucerdariancatalin | 8 | 421 | populating next right pointers in each node | 116 | 0.596 | Medium | 958 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1114796/python-recursion-constant-space-beats-98-short-solution | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root or not root.left:
return root
root.left.next = root.right
if root.next:
root.right.next = root.next.left
self.connect(root.left)
self.connect(root.right)
... | populating-next-right-pointers-in-each-node | python recursion constant space beats 98% short solution | marriema | 5 | 261 | populating next right pointers in each node | 116 | 0.596 | Medium | 959 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1786038/Python-3-(70ms)-or-BFS-O(N)-Solution | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
head = root
while root:
cur, root = root, root.left
while cur:
if cur.left:
cur.left.next = cur.right
if cur.next: cur.right.next = cur.next.... | populating-next-right-pointers-in-each-node | Python 3 (70ms) | BFS O(N) Solution | MrShobhit | 3 | 198 | populating next right pointers in each node | 116 | 0.596 | Medium | 960 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/726359/Python-Easy-understanding-solution-with-explaination | class Solution:
def connect(self, root: 'Node') -> 'Node':
# dfs
if not root:
return
# if root has non-empty right pointer, it means root must have left pointer and we need to link left -> right through next pointer
if root.right:
root.left.next = root.right
... | populating-next-right-pointers-in-each-node | Python -- Easy understanding solution with explaination | nbismoi | 3 | 177 | populating next right pointers in each node | 116 | 0.596 | Medium | 961 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/694695/Python3-O(1)-space-solution | class Solution:
def connect(self, root: 'Node') -> 'Node':
head = root
while head and head.left:
node = head
while node:
node.left.next = node.right
if node.next: node.right.next = node.next.left
node = node.next
h... | populating-next-right-pointers-in-each-node | [Python3] O(1) space solution | ye15 | 3 | 96 | populating next right pointers in each node | 116 | 0.596 | Medium | 962 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/694695/Python3-O(1)-space-solution | class Solution:
def connect(self, root: 'Node') -> 'Node':
def fn(node):
"""Connect node's children"""
if node and node.left:
node.left.next = node.right
if node.next: node.right.next = node.next.left
fn(node.left) or fn(node.... | populating-next-right-pointers-in-each-node | [Python3] O(1) space solution | ye15 | 3 | 96 | populating next right pointers in each node | 116 | 0.596 | Medium | 963 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/694695/Python3-O(1)-space-solution | class Solution:
def connect(self, root: 'Node') -> 'Node':
def fn(node, i=0):
"""Return tree whose next pointer is populated."""
if not node: return
node.next = seen.get(i)
seen[i] = node
fn(node.right, i+1) and fn(node.left, i+1)
... | populating-next-right-pointers-in-each-node | [Python3] O(1) space solution | ye15 | 3 | 96 | populating next right pointers in each node | 116 | 0.596 | Medium | 964 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/694695/Python3-O(1)-space-solution | class Solution:
def connect(self, root: 'Node') -> 'Node':
stack = [(0, root)]
seen = {}
while stack:
i, node = stack.pop()
if node:
node.next = seen.get(i)
seen[i] = node
stack.append((i+1, node.left))
... | populating-next-right-pointers-in-each-node | [Python3] O(1) space solution | ye15 | 3 | 96 | populating next right pointers in each node | 116 | 0.596 | Medium | 965 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/694695/Python3-O(1)-space-solution | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root: return # edge case
seen = {}
stack = [(0, root)]
while stack:
i, node = stack.pop()
if i in seen: seen[i].next = node
seen[i] = node
if node.right: stack.append... | populating-next-right-pointers-in-each-node | [Python3] O(1) space solution | ye15 | 3 | 96 | populating next right pointers in each node | 116 | 0.596 | Medium | 966 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/694695/Python3-O(1)-space-solution | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root: return # edge case
queue = [root]
while queue:
newq = []
next = None
for node in queue:
node.next = next
next = node
if node.right: ... | populating-next-right-pointers-in-each-node | [Python3] O(1) space solution | ye15 | 3 | 96 | populating next right pointers in each node | 116 | 0.596 | Medium | 967 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1102368/PythonPython3-Populating-Next-Right-Pointers-in-Each-Node | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root: return None
nodes_at_each_level = []
level = [root]
while level:
nodes_at_each_level.append([node for node in level])
level = [child for node in level for child ... | populating-next-right-pointers-in-each-node | [Python/Python3] Populating Next Right Pointers in Each Node | newborncoder | 2 | 195 | populating next right pointers in each node | 116 | 0.596 | Medium | 968 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1102368/PythonPython3-Populating-Next-Right-Pointers-in-Each-Node | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
def rp(parent, child):
if parent and child:
if child == parent.left:
child.next = parent.right
elif child == parent.right and parent.next:
... | populating-next-right-pointers-in-each-node | [Python/Python3] Populating Next Right Pointers in Each Node | newborncoder | 2 | 195 | populating next right pointers in each node | 116 | 0.596 | Medium | 969 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2391618/Python3-or-SC-O(1)-or-98-Faster-74-less-space-or-Easy-understanding | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
#if tree is empty
if not root:
return
#if tree has only one value
if not root.left or not root.right:
return root
flag = False #to assign the very first node.next = None
... | populating-next-right-pointers-in-each-node | [Python3] | SC O(1) | 98% Faster 74% less space | Easy-understanding | _vikaskumar_ | 1 | 56 | populating next right pointers in each node | 116 | 0.596 | Medium | 970 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2337682/Python-Simple-Iterative-solution-Space-O(1)-oror-Documented | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
parent = root
nextChild = root.left if root else None
# traverse the tree parent and next child that will be
# required once right side nodes are covered and we come back to this state
while pare... | populating-next-right-pointers-in-each-node | [Python] Simple Iterative solution - Space O(1) || Documented | Buntynara | 1 | 22 | populating next right pointers in each node | 116 | 0.596 | Medium | 971 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1655076/Python3-Intuitive-Recursive-Approach | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if not root: return
if root.next:
if root.right and root.next.left:
root.right.next = root.next.left
if root.left and root.right:
root.left.next = root.right
self.co... | populating-next-right-pointers-in-each-node | [Python3] Intuitive Recursive Approach | Rainyforest | 1 | 22 | populating next right pointers in each node | 116 | 0.596 | Medium | 972 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1653808/Python3-ITERATIVE-BFS-Explained | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if not root:
return
level = deque([root])
while level:
nextLevel = deque()
next = None
while level:
node = level.p... | populating-next-right-pointers-in-each-node | βοΈ [Python3] ITERATIVE BFS, Explained | artod | 1 | 114 | populating next right pointers in each node | 116 | 0.596 | Medium | 973 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1653748/Python3-2-Solutions-or-BFS-O(n)-and-Modified-BFS-O(1) | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if not root: return None
level = [root]
while level:
level.append(None)
nextLevel = []
for i in range(len(level) - 1):
level[i].next = level[i + 1]
... | populating-next-right-pointers-in-each-node | [Python3] 2 Solutions | BFS O(n) & Modified BFS O(1) | PatrickOweijane | 1 | 106 | populating next right pointers in each node | 116 | 0.596 | Medium | 974 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1653748/Python3-2-Solutions-or-BFS-O(n)-and-Modified-BFS-O(1) | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if not root: return None
level = root
while level:
nextLevel = level.left
last = None
while level:
if last: last.next = level.left
if level.left:... | populating-next-right-pointers-in-each-node | [Python3] 2 Solutions | BFS O(n) & Modified BFS O(1) | PatrickOweijane | 1 | 106 | populating next right pointers in each node | 116 | 0.596 | Medium | 975 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1595860/3-iterative-solutions-in-Python-from-O(n)-space-to-O(1) | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
from collections import deque
q = deque([(root, 0)])
while q:
n, level = q.popleft()
q += ((c, level + 1) for c in (n.left, n.right) if c)
try:
... | populating-next-right-pointers-in-each-node | 3 iterative solutions in Python, from O(n) space to O(1) | mousun224 | 1 | 145 | populating next right pointers in each node | 116 | 0.596 | Medium | 976 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1595860/3-iterative-solutions-in-Python-from-O(n)-space-to-O(1) | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
level = [root]
while level:
for i in range(len(level) - 1):
level[i].next = level[i + 1]
level = [c for n in level for c in (n.left, n.right) if c]
... | populating-next-right-pointers-in-each-node | 3 iterative solutions in Python, from O(n) space to O(1) | mousun224 | 1 | 145 | populating next right pointers in each node | 116 | 0.596 | Medium | 977 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1595860/3-iterative-solutions-in-Python-from-O(n)-space-to-O(1) | class Solution:
def connect(self, root: 'Node') -> 'Node':
head = root
while head and head.left:
left_most = head.left
while head:
head.left.next = head.right
if head.next:
head.right.next = head.next.left
he... | populating-next-right-pointers-in-each-node | 3 iterative solutions in Python, from O(n) space to O(1) | mousun224 | 1 | 145 | populating next right pointers in each node | 116 | 0.596 | Medium | 978 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1572626/Python-or-Recursive-approach-with-no-extra-space-or-Explained | class Solution:
def connect(self, root: 'Node') -> 'Node':
# corner case where the tree is empty
if root:
# if it's not a leaf node
if root.left and root.right:
# connect the left child to the right child
root.left.next = root.right
... | populating-next-right-pointers-in-each-node | Python | Recursive approach with no extra space | Explained | hemersontacon | 1 | 73 | populating next right pointers in each node | 116 | 0.596 | Medium | 979 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1478239/Python-O(1)-space-with-recursion-beats-90-speed | class Solution:
def connect(self, root: 'Node') -> 'Node':
if root is None:
return root
self.traverse(root, None, None)
return root
def traverse(self, node, parent, goLeft):
if node is None:
return
left, right = node.left, n... | populating-next-right-pointers-in-each-node | Python O(1) space with recursion beats 90 speed | SleeplessChallenger | 1 | 170 | populating next right pointers in each node | 116 | 0.596 | Medium | 980 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1432505/Row-by-row-80-speed | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root
row = [root]
while row:
new_row = []
for i, node in enumerate(row):
if i < len(row) - 1:
node.next = row[i + 1]
if node.... | populating-next-right-pointers-in-each-node | Row by row, 80% speed | EvgenySH | 1 | 82 | populating next right pointers in each node | 116 | 0.596 | Medium | 981 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2842198/Python-DFS | class Solution:
def dfs(self, root, levels, level):
if level not in levels:
levels[level] = []
if root and root.left:
levels[level].extend([root.left, root.right])
self.dfs(root.left, levels, level + 1)
self.dfs(root.right, levels, level + 1)
de... | populating-next-right-pointers-in-each-node | Python DFS | alimohammad1995 | 0 | 1 | populating next right pointers in each node | 116 | 0.596 | Medium | 982 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2677789/Python3-or-Hash-Map-%2B-BFS | class Solution:
def __init__(self):
self.hash_table = None
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if root is None:
return
self.hash_table = defaultdict(list)
self.level_order(root, 0)
for key in self.hash_table:
if key == 0... | populating-next-right-pointers-in-each-node | Python3 | Hash Map + BFS | honeybadgerofdoom | 0 | 6 | populating next right pointers in each node | 116 | 0.596 | Medium | 983 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2657457/Python3-DFS-recursion!-Nodes-matched-on-the-same-level | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
tree = []
def dfs(node, level=0):
if node is None:
return
if len(tree) == level:
tree.append([])
dfs(node.right, level + 1)
dfs(... | populating-next-right-pointers-in-each-node | Python3 DFS, recursion! Nodes matched on the same `level` | zakmatt | 0 | 31 | populating next right pointers in each node | 116 | 0.596 | Medium | 984 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2610043/Python3-faster-than-99.77-submissions | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
curr , nxt = root, root.left if root else None
while curr and nxt:
curr.left.next = curr.right
if curr.next:
curr.right.next = curr.next.left
curr = curr.next
... | populating-next-right-pointers-in-each-node | Python3 faster than 99.77% submissions | sumedha19129 | 0 | 25 | populating next right pointers in each node | 116 | 0.596 | Medium | 985 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2469191/simple-python-solution-using-level-order-and-bfs-with-comments | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if(not root):
return None
q = deque()
q.append(root)
while(q):
sz = len(q)
pre = None
for i in range(sz):
curr = q.popleft()
... | populating-next-right-pointers-in-each-node | simple python solution using level order and bfs with comments | rajitkumarchauhan99 | 0 | 17 | populating next right pointers in each node | 116 | 0.596 | Medium | 986 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2283706/116.-My-Java-and-Python-Solution-with-comments | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if not root:
return None
dummy = root
# while we still have next level to check
while root.left:
# save the pointer of next level
next_level = root.left
... | populating-next-right-pointers-in-each-node | 116. My Java and Python Solution with comments | JunyiLin | 0 | 26 | populating next right pointers in each node | 116 | 0.596 | Medium | 987 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2220136/Python-recursive...Very-simple-and-commented | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
def dfs(node, nxt):
if not node:
return
node.next = nxt
dfs(node.left, node.right) # left and right children connect with next
if node.next:
... | populating-next-right-pointers-in-each-node | Python recursive...Very simple and commented | saqibmubarak | 0 | 46 | populating next right pointers in each node | 116 | 0.596 | Medium | 988 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2149548/Python3-Solution | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if not root:
return None
self.traverse(root.left, root.right);
return root
def traverse(self, node1, node2):
if not node1 or not node2:
return
node1.next = no... | populating-next-right-pointers-in-each-node | Python3 Solution | qywang | 0 | 38 | populating next right pointers in each node | 116 | 0.596 | Medium | 989 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/2061341/Python-level-order-traversal | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
q = deque([root]) if root else None
while q:
length = len(q)
prev = None
for _ in range(length):
node = q.popleft()
node.next = prev
prev... | populating-next-right-pointers-in-each-node | Python, level order traversal | blue_sky5 | 0 | 18 | populating next right pointers in each node | 116 | 0.596 | Medium | 990 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1976150/Python-DFS-oror-O(N)-time-and-O(1)-Memory-Complexity | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
cur, nextNode = root, root.left if root else None
while cur and nextNode:
cur.left.next = cur.right
if cur.next:
cur.right.next = cur.next.left
cur... | populating-next-right-pointers-in-each-node | Python - DFS || O(N) time and O(1) Memory Complexity | dayaniravi123 | 0 | 70 | populating next right pointers in each node | 116 | 0.596 | Medium | 991 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1864709/Python3-Simple-recursive-solution | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if (root == None): return None
self.connectTwoNodes(root.left, root.right)
return root
def connectTwoNodes(self, node1, node2):
if (node1 == None or node2 == None): return
... | populating-next-right-pointers-in-each-node | [Python3] Simple recursive solution | leqinancy | 0 | 22 | populating next right pointers in each node | 116 | 0.596 | Medium | 992 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1804220/simple-python-bfs | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
cur = [root]
nxt = []
while root and len(cur) > 0:
cur.append(None)
for i in range(0, len(cur)-1):
cur[i].next = cur[i+1]
if cur[i].left: nxt.append(cur[i].l... | populating-next-right-pointers-in-each-node | simple python bfs | gasohel336 | 0 | 45 | populating next right pointers in each node | 116 | 0.596 | Medium | 993 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1785696/Easy-to-Understand-or-Step-by-Step-Explanation | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
q1 = []
q1.append(root)
level = 0
if not root:
return root
while q1:
rightMostNodeIndex = 2**level - 1
q1[rightMostNodeIndex].next = None
for i in ra... | populating-next-right-pointers-in-each-node | Easy to Understand | Step by Step Explanation | swap2210 | 0 | 26 | populating next right pointers in each node | 116 | 0.596 | Medium | 994 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node/discuss/1763650/Python-Simple-Python-Solution-Using-BFS-With-the-help-of-Queue | class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
if root is None:
return None
queue = [root]
total_node = 1
while queue:
count_node = 1
for i in range(len(queue)):
node = queue.pop(0)
if count_node == total_node:
node.next = None
else:
node.n... | populating-next-right-pointers-in-each-node | [ Python ] ββ Simple Python Solution Using BFS With the help of Queue βπ₯π₯ | ASHOK_KUMAR_MEGHVANSHI | 0 | 122 | populating next right pointers in each node | 116 | 0.596 | Medium | 995 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/2033286/Python-Easy%3A-BFS-and-O(1)-Space-with-Explanation | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
q = deque()
q.append(root)
dummy=Node(-999) # to initialize with a not null prev
while q:
length=len(q) # find level length
prev=dummy
... | populating-next-right-pointers-in-each-node-ii | Python Easy: BFS and O(1) Space with Explanation | constantine786 | 39 | 3,000 | populating next right pointers in each node ii | 117 | 0.498 | Medium | 996 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/2033286/Python-Easy%3A-BFS-and-O(1)-Space-with-Explanation | class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return None
curr=root
dummy=Node(-999)
head=root
while head:
curr=head # initialize current level's head
prev=dummy # init prev for next level linke... | populating-next-right-pointers-in-each-node-ii | Python Easy: BFS and O(1) Space with Explanation | constantine786 | 39 | 3,000 | populating next right pointers in each node ii | 117 | 0.498 | Medium | 997 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/521193/Python-O(1)-aux-space-DFS-sol.-with-Diagram | class Solution:
def connect(self, root: 'Node') -> 'Node':
def helper( node: 'Node'):
if not node:
return None
scanner = node.next
# Scanner finds left-most neighbor, on same level, in right hand side
while s... | populating-next-right-pointers-in-each-node-ii | Python O(1) aux space DFS sol. [with Diagram] | brianchiang_tw | 31 | 1,500 | populating next right pointers in each node ii | 117 | 0.498 | Medium | 998 |
https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/discuss/1448257/Python3-in-place-BFS-solution-faster-than-97 | class Solution:
def connect(self, root: 'Node') -> 'Node':
parents = [root]
kids = []
prev = None
while len(parents) > 0:
p = parents.pop(0)
if prev:
prev.next = p
prev = p
if p:
if p.left:
... | populating-next-right-pointers-in-each-node-ii | Python3 - in-place, BFS solution, faster than 97% | elainefaith0314 | 4 | 255 | populating next right pointers in each node ii | 117 | 0.498 | Medium | 999 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.