[leetcode] 337.House Robber III

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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     }

[leetcode] 337.House Robber III

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原文地址：http://www.cnblogs.com/ym65536/p/5616772.html