23. House Robber III.cpp

/*
House Robber III
================

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.
*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution
{
  pair<int, int> traverse(TreeNode *root)
  {
    if (!root)
      return {0, 0};
    auto left = traverse(root->left);
    auto right = traverse(root->right);
    int NotRob = max(left.first + right.first, max(left.first + right.second, max(left.second + right.first, left.second + right.second)));

    return {root->val + left.second + right.second, NotRob};
  }

public:
  int rob(TreeNode *root)
  {
    auto ans = traverse(root);
    return max(ans.first, ans.second);
  }
};