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.
1 struct Money { 2 int pre; 3 int curr; 4 Money():pre(0), curr(0){} 5 }; 6 7 int rob(TreeNode* root) { 8 Money sum = dfs(root); 9 return sum.curr; 10 } 11 12 Money dfs(TreeNode* root) 13 { 14 if (root == NULL) return Money(); 15 Money leftMoney = dfs(root->left); 16 Money rightMoney = dfs(root->right); 17 Money sumMoney; 18 sumMoney.pre = leftMoney.curr + rightMoney.curr; // 当前节点不偷 19 sumMoney.curr = max(sumMoney.pre, root->val + leftMoney.pre + rightMoney.pre); 20 return sumMoney; 21 }
时间: 2024-10-10 16:53:35