[leetcode] 337.House Robber III

Posted

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[leetcode] 337.House Robber III相关的知识,希望对你有一定的参考价值。

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.

 1 struct Money {
 2         int pre;  
 3         int curr; 
 4         Money():pre(0), curr(0){}
 5     };
 6 
 7     int rob(TreeNode* root) {
 8         Money sum = dfs(root);
 9         return sum.curr;
10     }
11 
12     Money dfs(TreeNode* root)
13     {
14         if (root == NULL) return Money();
15         Money leftMoney = dfs(root->left);   
16         Money rightMoney = dfs(root->right); 
17         Money sumMoney;
18         sumMoney.pre = leftMoney.curr + rightMoney.curr; // 当前节点不偷
19         sumMoney.curr = max(sumMoney.pre, root->val + leftMoney.pre + rightMoney.pre);
20         return sumMoney;
21     }

 

以上是关于[leetcode] 337.House Robber III的主要内容,如果未能解决你的问题,请参考以下文章

[leetcode] 337.House Robber III

leetcode No337. House Robber III

<LeetCode OJ> 337. House Robber III

LeetCode in Python 337. House Robber III

leetcode 337. House Robber III

LeetCode 337. House Robber III 动态演示