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
3
/ \
2 3
\ \
3 1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
3
/ \
4 5
/ \ \
1 3 1
Maximum amount of money the thief can rob = 4 + 5 = 9.
LeetCode上的原题，请参见我之前的博客House Robber III 。
解法一：
class Solution { public: int houseRobber3(TreeNode* root) { unordered_map<TreeNode*, int> m; return helper(root, m); } int helper(TreeNode *root, unordered_map<TreeNode*, int> &m) { if (!root) return 0; if (m.count(root)) return m[root]; int val = 0; if (root->left) { val += helper(root->left->left, m) + helper(root->left->right, m); } if (root->right) { val += helper(root->right->left, m) + helper(root->right->right, m); } val = max(val + root->val, helper(root->left, m) + helper(root->right, m)); m[root] = val; return val; } };
解法二：
class Solution { public: int houseRobber3(TreeNode* root) { vector<int> res = helper(root); return max(res[0], res[1]); } vector<int> helper(TreeNode *root) { if (!root) return {0, 0}; vector<int> left = helper(root->left); vector<int> right = helper(root->right); vector<int> res{0, 0}; res[0] = max(left[0], left[1]) + max(right[0], right[1]); res[1] = left[0] + right[0] + root->val; return res; } };
本文转自博客园Grandyang的博客，原文链接：打家劫舍之三[LintCode] House Robber III ，如需转载请自行联系原博主。
