题解 | #二叉树中和为某一值的路径(一)#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @param sum int整型 
# @return bool布尔型
#
class Solution:
    def hasPathSum(self , root: TreeNode, sum: int) -> bool:
        # write code here
        def dfs(root,sum):
            if not root:
                return False
            if root.val == sum and root.left is None and root.right is None:
                return root
            return dfs(root.left,sum-root.val) or dfs(root.right,sum-root.val)
        return dfs(root,sum)