solve

Problem

DFS, simple question: only check when we are at a leaf node (somehow my solution was top 99% in terms of both time and space complexity)

from typing import Optional
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        return self.dfs(root, currSum=0, targetSum=targetSum)

    def dfs(self, root, currSum, targetSum) -> bool:

        if not root:
            return False
        
        if not root.left and not root.right:
            if root.val + currSum == targetSum:
                return True

        else:
            currSum = root.val + currSum
            return self.dfs(root.left, currSum, targetSum) or self.dfs(root.right, currSum, targetSum)