[Leetcode]112. Path Sum(C++)

题目描述

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.

例子

Input: [5,4,8,11,null,13,4,7,2,null,null,null,1] Output: 22 Explanation: there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Note

A leaf is a node with no children.

解题思路

采用递归方法，执行以下步骤：

当 root 为空，返回 false
当 root 为叶结点且其值等于 sum 返回 true
对其左右节点递归调用 hasPathSum()，将 sum 设为 sum - root->val，其中一棵子树返回 true 即返回 true

代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (!root) return false;
        if (!root->left && !root->right && root->val == sum) return true;
        return hasPathSum(root->left, sum - root->val) ||
               hasPathSum(root->right, sum - root->val);
    }
};

时间复杂度: O(n)
空间复杂度: O(h) 递归深度

GitHub 代码同步地址： 112.PathSum.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions