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   1Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.Example 2:3/ \4   5/ \   \1   3   1Maximum 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);}};