Path Sum 2 in Binary Tree | Leetcode 113 Solution

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Explained very well! Was able to code it on my own without seeing the video further :)

ayushidalal

Good explanation of why we have to create a new list every time (Shallow cloning concept).

chhaviagarwal

i have also used the same approach, but the same code fails for nodes with negative values, can u please confirm whether this code runs for -ve values as well ?

DurgaShiva

What will be the time and space complexity?

manantyagi

Nice video. Please do a video on path sum 3 also with complexities.

pavantejak

Adding(line - 27) and removing(line - 32) from smallAns list is redundant. Without that also code will work if we pass the updated list in function call.

class Solution:
def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
def helper(root, targetSum, path):
if not root:
return
if root.left is None and root.right is None:
if targetSum - root.val == 0:
self.answer.append(path + [root.val])
helper(root.left, targetSum - root.val, path + [root.val])
helper(root.right, targetSum - root.val, path + [root.val])
self.answer = []
helper(root, targetSum, [])
return self.answer

TheShantanu

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