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.

样例

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \ 
 1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

算法1

(dfs)

全局最优+局部最优

unordered_map<TreeNode*,int> map相当于记录最优解，
又是一个记忆化搜索的问题

c++代码

class Solution {
public:
    int rob(TreeNode* root) {
        unordered_map<TreeNode*,int> m;
        return dfs(root,m);
    }
    int dfs(TreeNode *root,unordered_map<TreeNode*,int>&m){
        if(root==nullptr)
            return 0;
        if(m.find(root)!=m.end())
            return m[root];
        int val=0;
        if(root->left){
            val+=dfs(root->left->left,m)+dfs(root->left->right,m);
        }
        if(root->right){
            val+=dfs(root->right->left,m)+dfs(root->right->right,m);
        }
        val=max(root->val+val,dfs(root->left,m)+dfs(root->right,m));
        m[root]=val;
        return val;
    }
};