LeetCode No.337. House Robber III

1. 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.

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.

2. 思路

题目说明提到不能同时偷窃相邻的两间房间。
那么最大的偷窃为 选取根节点的， 不选取根节点的。
详细请参见源代码。

3. 代码及复杂度分析

class Solution {public:    int rob(TreeNode* root) {        vector<int> whole_rob = CurrRob(root);        return max(whole_rob[0], whole_rob[1]);    }private:    vector<int> CurrRob(TreeNode * root) {        vector<int> ret(2, 0);        if (root == nullptr) return ret;        vector<int> left_rob = CurrRob(root->left);        vector<int> right_rob = CurrRob(root->right);        ret[0] = max(left_rob[0], left_rob[1]) + max(right_rob[0], right_rob[1]); // without rob root        ret[1] = root->val + left_rob[0] + right_rob[0];// rob root         return ret;    }};

4. 参考资料

1. https://leetcode.com/discuss/91899/step-by-step-tackling-of-the-problem