动态规划解337. House Robber III

2024年6月10日 131次阅读

题目

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.

Example 1:

 3
/ \
2  3
\   \ 
 3   1

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

  3
 / \
 4   5
/ \  \
1 3   1

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

题意分析

一个盗贼进入一个居民区，它发现只有一个根可以进入该居民区。除了这个根，其他的居民家有且只有一个连着它的父亲居民家。盗贼经过侦查之后，发现这个居民区的组成跟二叉树一样，并且如果两个直连的居民家在同一晚上被偷，则会引起警铃，现在要求盗贼能偷的最大价值

思路分析

这是一道动态规划题目，对于每个节点，用2个变量去表示，分别为result[0],result[1]，result[0]表示不选择当前节点子树的和，result[1]表示选择当前节点子树的和，用leftvlaue表示当前节点左子树的节点变量，用rightvalue表示当前节点右子树的节点变量，则result[0]=max(leftvalue[0], leftvalue[1]) + max(rightvalue[0], rightvalue[1])， result[1]= root->val + leftvalue[1], + rightvalue[1],这个过程通过dfs深度递归完成

AC代码

class Solution {
public:
    vector<int> dfs(TreeNode* subroot) {
        vector<int> result(2, 0);
        if (subroot == nullptr) return result;
        vector<int> leftvalue = dfs(subroot->left);
        vector<int> rightvalue = dfs(subroot->right);
        result[0] = max(leftvalue[0], leftvalue[1]) + max(rightvalue[0], rightvalue[1]);
        result[1] = subroot->val + leftvalue[0] + rightvalue[0];
        return result;
    }
    int rob(TreeNode* root) {
        vector<int> res = dfs(root);
        return max(res[0], res[1]);
    }
};
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