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)
原文：https://www.cnblogs.com/boluo007/p/12434451.html