path sum 有好几题，是一个系列。
原题是：
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.
思路是：
递归函数，上一层和下一层的沟通通道，如果是从上到下，通过传入参数；如果是从下到上，通过函数返回值。
我们把从root开始的节点的和，sumPath，作为参数，一路传递下去，在函数中，直接和目标值target相比较，相同就返回True, 不相同就返回False.
同时做这个比较判断时，当前这个节点需要为叶子节点。否则，继续迭代到下一层。
代码是：
# 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, target):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if root == None:
return False
else:
return self.helper(root,0,target)
def helper(self,root,sumPath,target):
sumPath = sumPath + root.val
if not(root.left or root.right):
if sumPath == target:
return True
else:
if root.left and self.helper(root.left, sumPath,target):
return True
if root.right and self.helper(root.right, sumPath,target):
return True
return False
