LeetCode No.337. House Robber III

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本文介绍了一道经典的二叉树遍历问题——打家劫舍。文章详细阐述了如何通过递归的方式求解最大劫财数额，并给出具体代码实现。通过选择是否抢劫根节点对应的房屋来决定后续的策略，确保相邻房屋不会同时被劫。

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:

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:

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;
    }
};