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.
Tags: Depth-first Search
分析
深度优先遍历题目，拿到题后首先需要确认几个题中没有明确给出的要点：
- root-to-leaf 路径，必须是从根结点一直到叶子结点，中间取一段是不行的
- 结点值是否可以为负。如果不可以为负，那么在判断中一旦发现从根结点到当前结点的和大于期望的值，就可以直接短路返回否。但是本题中结点值可以为负
- 空的二叉树，是否可以认为存在和为0的路径。本题中不可以
这些要点判断出来后，剩余的工作就是构建递归的深度优先遍历，在每层判断是否是叶子结点、是否存在加和等于期望值的路径。
示例
class Solution:
# @param root, a tree node
# @param sum, an integer
# @return a boolean
def hasPathSum(self, root, sum):
if root is None:
return False
else:
# Only leaf could return true
if sum == root.val and root.left is None and root.right is None:
return True
else:
return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
Leetcode 笔记系列的Python代码共享在https://github.com/wizcabbit/leetcode.solution