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

#
# 
# @param root TreeNode类 
# @param sum int整型 
# @return bool布尔型
#
class Solution:
    def hasPathSum(self , root , sum ):
        # write code here
        if not root:
            return False
        if sum==root.val and not root.left and not root.right:
            return True
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)