[leetcode]337. House Robber III

Analysis

fighting!!!—— [每天刷題並不難0.0]

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.

Implement

/**
 * 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) {
        vector<int> res = DFS(root);
        return max(res[0], res[1]);
    }
    vector<int> DFS(TreeNode* node){
        vector<int> res(2, 0);
        if(!node)
            return res;
        vector<int> left = DFS(
 
node->left);
        vector<int> right = DFS(node->right);
        res[0] = max(left[0], left[1])+max(right[0], right[1]);
        res[1] = left[0]+right[0]+node->val;
        return res;
    }
};