337. House Robber III

• Difficulty: Medium

• Topics: Tree, Depth-first Search

• Similar Questions:

  • House Robber

  • House Robber II

Problem:

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. </pre>

Solutions:

/**
 * 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 {
public:
    int rob(TreeNode* root) {
        return rob(root, true);

    }

private:
    int rob(TreeNode* root, bool couldRob) {
        if (root == nullptr)    return 0;
        if (cache.count({root, couldRob})) {
            return cache[{root, couldRob}];
        }

        int robRoot = 0;
        if (couldRob) {
            robRoot = root->val + rob(root->left, false) + rob(root->right, false);
        }

        int notRobRoot = 0;
        notRobRoot = rob(root->left, true) +  rob(root->right, true);

        cache[{root, couldRob}] = max(robRoot, notRobRoot);
        return max(robRoot, notRobRoot);
    }

    map<pair<TreeNode*, bool>, int> cache;

};