112. Path Sum

原文题目：

112. Path Sum

读题：

二叉树中是否存在给定和为给定数的路径，递归实现最简单

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def hasPathSum(self, root, sum):        """        :type root: TreeNode        :type sum: int        :rtype: bool        """        if not root:            return False        if not root.left and not root.right and sum == root.val:            return True        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)