【LEETCODE】112-Path Sum

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Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:

Given the below binary tree and sum = 22,

/ \

4 8

/ / \

11 13 4

/ \ \

7 2 1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

题意：

给一个二叉树和一个和，判断是否存在一个从root到leaf的path，使得其中的所有值的和等于sum

思路：

遍历每个path，计算sum，直到等于22或者遍历结束

或者遇到一个点，sum减去节点的值，到leaf时剩余0则True，否则False

参考：

http://www.cnblogs.com/CheeseZH/p/4034291.html

# 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 root is None:            return False                sum-=root.val              #遇到一个点，sum减去节点的值                        if sum==0 and root.left==None and root.right==None:         #到leaf时剩余0则True，否则False            r=True        else:            r=False                return r or self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)      #<span style="font-family: Arial, Helvetica, sans-serif;">直到leaf</span>

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