Leetcode 337. House Robber III (Medium) (cpp)

Leetcode 337. House Robber III (Medium) (cpp)

Tag: Tree, Depth-first Search

Difficulty: Medium

/*

337. House Robber III (Medium)

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:
3
/ \
2   3
\   \
3   1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3
/ \
4   5
/ \   \
1   3   1
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(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
	int rob(TreeNode* root) {
		int l = 0, r = 0;
		return rob(root, l, r);
	}
private:
	int rob(TreeNode* root, int& l, int& r) {
		if (root == NULL) {
			return 0;
		}
		int ll = 0, lr = 0, rl = 0, rr = 0;
		l = rob(root->left, ll, lr);
		r = rob(root->right, rl, rr);
		return max(root->val + ll + lr + rl + rr, l + r);
	}
};