LintCode 535: House Robber III

535. House Robber III
     中文English
     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
Example1

Input: {3,2,3,#,3,#,1}
Output: 7
Explanation:
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
3
/
2 3
\ \
3 1
Example2

Input: {3,4,5,1,3,#,1}
Output: 9
Explanation:
Maximum amount of money the thief can rob = 4 + 5 = 9.
3
/
4 5
/ \ \
1 3 1
Notice
This problem is the extention of House Robber and House Robber II

解法1：DFS
每个maxChildTree()都考虑爷孙组合和左右儿子组合，取其优。
代码如下:

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: The maximum amount of money you can rob tonight
     */
    int houseRobber3(TreeNode * root) {
        return maxChildTree(root);
    }

private:
    int maxChildTree(TreeNode * root) {
        
        if (!root) return 0;
        if (!root->left && !root->right) return max(0, root->val);
        if (!root->left) return max(maxChildTree(root->right->left) + maxChildTree(root->right->right) + max(0, root->val), maxChildTree(root->right));
        if (!root->right) return max(maxChildTree(root->left->left) + maxChildTree(root->left->right) + max(0, root->val), maxChildTree(root->left));
        
        int option1 = maxChildTree(root->left->left) + maxChildTree(root->left->right) + maxChildTree(root->right->left) + maxChildTree(root->right->right) + max(0, root->val);
        
        int option2 = maxChildTree(root->left) + maxChildTree(root->right);
        
        return max(option1, option2);
    }
};