112. Path Sum刷题笔记

简单的递归就能做

# 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:
        if not root:
            return False
        if not root.left and not root.right:
            if root.val == targetSum:
                return True
            else:
                return False
        return self.hasPathSum(root.left,targetSum - root.val) or self.hasPathSum(root.right,targetSum - root.val)
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