【leetcode】337. House Robber III

題目：
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.

---

思路：
抄作業https://www.cnblogs.com/grandyang/p/5275096.html，引用一下關鍵內容：

遞歸函數返回一個大小爲2的一維數組 res，其中 res[0] 表示不包含當前節點值的最大值，res[1] 表示包含當前值的最大值。

---

代碼實現：

/**
 * 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:
    
    vector<int> f(TreeNode *root){
        if (root == nullptr){
            return {0,0};
        }
        
        vector<int> left_ret = f(root->left);
        vector<int> right_ret = f(root->right);
        vector<int> ret(2,0);
        ret[0] = max(left_ret[0], left_ret[1]) + max(right_ret[0], right_ret[1]);
        ret[1] = left_ret[0] + right_ret[0] + root->val;
        
        return ret;
    }
    
    int rob(TreeNode* root) {
        if (root == nullptr){
            return 0;
        }
        
        vector<int> ret = f(root);
        return max(ret[0], ret[1]);
    }
};

---

參考：
https://www.cnblogs.com/grandyang/p/5275096.html