# 337. 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.

# Solution

Approach 1: DFS -- need to distinguish between situations where the current node is robbed or not.

# Code (Python)

Approach 1:

    def rob(self, root: TreeNode) -> int:
        return max(self._rob(root))
    
    def _rob(self, node):
        # returns a tuple of (max_rob_if_rob_current_node, max_rob_if_not_rob_current_node)
        if not node:
            return (0, 0)
        if not node.left and not node.right:
            return (node.val, 0)
        left_rob, right_rob = self._rob(node.left), self._rob(node.right)
        not_rob_current = left_rob[0] + right_rob[0]
        # max of robbing current means you can either rob or not rob the current node
        # but max of not robbing the current node has to exclude the current node, otherwise we violate the rules
        rob_current = max(node.val + left_rob[1] + right_rob[1], not_rob_current)
        return (rob_current, not_rob_current)

# Code (C++)

Approach 1:

/**
 * 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 {
private:
    unordered_map<TreeNode*,int> umap;
public:
    int rob(TreeNode* root) {
        if (root == NULL)
            return 0;
        if (umap.find(root) != umap.end())
            return umap[root];
        int val = root->val;
        if (root->left)
            val += rob(root->left->left) + rob(root->left->right);
        if (root->right)
            val += rob(root->right->left) + rob(root->right->right);
        val = std::max(val, rob(root->left) + rob(root->right));
        umap[root] = val;
        return val;
    }
};