# 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;
}
};