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0337. House Robber III.md

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0337. House Robber III, medium, , freq: 11%, acceptance: 48.5%

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:

Input: [3,2,3,null,3,null,1]

 3
/ \

2 3 \ \ 3 1

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

Input: [3,4,5,1,3,null,1]

 3
/ \

4 5 / \ \ 1 3 1

Output: 9 Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

// 12ms, 93%
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    // return pair amount with & without robbing root
    pair<int,int> check(TreeNode* root) {
        if (!root) return {0, 0};
        auto l = check(root->left);
        auto r = check(root->right);
        return { root->val + l.second + r.second,
               max(l.first,l.second) + max(r.first,r.second)};
    }
public:
    int rob(TreeNode* root) {
        auto r = check(root);
        return max(r.first, r.second);
    }
};