Leetcode 笔记 112 - Path Sum

题目链接:Path Sum | LeetCode OJ

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

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