【LEETCODE】112-Path Sum

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,
5
/ \
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 = None

class 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>