337. House Robber III

1. Description

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.

2. Example

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.

3. Constraints

• The number of nodes in the tree is in the range [1, $10^4$].
• 0 <= Node.val <= $10^4$

4. Solutions

Dynamic Programming

n is the number of nodes in root
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    int rob(TreeNode *root) {
        auto robs = rob_(root);
        return max(robs[0], robs[1]);
    }

private:
    vector<int> rob_(TreeNode *root) {
        // 0 rob, 1 not rob
        if (root != nullptr) {
            auto left_rob = rob_(root->left);
            auto right_rob = rob_(root->right);

            int rob_profit = root->val + left_rob[1] + right_rob[1];
            int not_rob_profit = max(left_rob[0], left_rob[1]) + max(right_rob[0], right_rob[1]);

            return {rob_profit, not_rob_profit};
        } else {
            return {0, 0};
        }
    }
};