[LintCode] House Robber III 打家劫舍之三

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 ;
        if (m.count(root)) return m[root];
        int val = ;
        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[], res[]);
    }
    vector<int> helper(TreeNode *root) {
        if (!root) return {, };
        vector<int> left = helper(root->left);
        vector<int> right = helper(root->right);
        vector<int> res{, };
        res[] = max(left[], left[]) + max(right[], right[]);
        res[] = left[] + right[] + root->val;
        return res;
    }
};