LeetCode-337 House Robber III

题目描述

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

题目大意

给定一棵二叉树，不能同时选择相邻结点的数值，求可能得到的结点值之和的最大值。

示例

E1

E2

解题思路

递归遍历，计算当前结点以及其孙子结点的数值之和，和该结点的两个孩子结点之和进行比较，返回较大值。

复杂度分析

时间复杂度：O(N)

空间复杂度：O(N)

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int rob(TreeNode* root) {
        int l = 0, r = 0;
        
        return dfs(root, l, r);
    }
    
    int dfs(TreeNode* root, int& l, int& r) {
        if(root == NULL)
            return 0;
        
        int ll = 0, lr = 0, rl = 0, rr = 0;
        l = dfs(root->left, ll, lr);
        r = dfs(root->right, rl, rr);
        
        return max(root->val + ll + lr + rl + rr, l + r);
    }
};