[LeetCode] 337. House Robber III

[frdmu]

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called 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 form a binary tree. It will automatically contact the police if two directly-linked houses were broken into on the same night.

Given the root of the binary tree, return the maximum amount of money the thief can rob without alerting the police.

 

Example 1:

Input: root = [3,2,3,null,3,null,1]
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

Input: root = [3,4,5,1,3,null,1]
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• 0 <= Node.val <= 104

 

解法：
动态规划。思路和前两道一样。代码如下：

class Solution {
public:
    map<TreeNode*, int> mp;
    int rob(TreeNode* root) {
        if (!root) return 0;
        if (mp.find(root) != mp.end())
            return mp[root];
        
        int do_it = root->val + 
                    (root->left ? rob(root->left->left) + rob(root->left->right) : 0) +
                    (root->right? rob(root->right->left) + rob(root->right->right) : 0);
        int undo = rob(root->left) + rob(root->right);
        int res = max(do_it, undo);
        mp[root] = res;
        
        return res; 
    }
};

Refer:
团灭 LeetCode 打家劫舍问题