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.

Note: A leaf is a node with no children.

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.

# 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
        """
        
        total = 0
        return self.helper(root, sum, total)
    
    def helper(self, root, sum, total):
        #只有单个子节点的非耶节点的情况
        if root is None:
            return False
        #叶节点结算值，返回结果
        if root.left is None and root.right is None:
            total += root.val
            if total == sum:
                return True
            else:
                return False
        total += root.val
        
        return self.helper(root.left, sum, total) or self.helper(root.right, sum, total)