[easy+][Tree]112. Path Sum

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