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
# 返回型递归
class Solution:
    def hasPathSum(self , root: TreeNode, sum: int) -> bool:
        if root is None:
            return False
        if not root.left and not root.right and sum-root.val==0:
            return True
        return self.hasPathSum(root.right, sum-root.val) or self.hasPathSum(root.left, sum-root.val)