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.
vector<int> dfs(TreeNode *root){ vector<int> res = { 0, 0 }; //res[0]表示抢，res[1]表示不抢 if (root == NULL)return res; vector<int> lres = dfs(root->left); vector<int> rres = dfs(root->right); res[0] = lres[1] + rres[1] + root->val;//抢当前节点，那么左右孩子不能抢 res[1] = max(lres[0], lres[1]) + max(rres[0], rres[1]);//不抢当前节点，则左右孩子可抢可不抢 return res; } int rob(TreeNode* root) { vector<int> res = dfs(root); return max(res[0], res[1]); }