[leetcode-337-House Robber III]

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.

思路：

Basically you want to compare which one is bigger between 1) you + sum of your grandchildren and 2) sum of your children. 

用l表示下一层左子树的值，r表示下一层右子树的值。那么对于root的值只需比较

max(l + r, root->val + ll + lr + rl + rr)即可。

int rob3(TreeNode* root, int& l, int& r)
    {
        if (root == NULL)return 0;
        int ll = 0, lr = 0, rl = 0, rr = 0;
        l = rob3(root->left, ll, lr);
        r = rob3(root->right, rl, rr);
        return max(l + r, root->val + ll + lr + rl + rr);
    }
    int rob3(TreeNode* root)
    {
        int l, r;
        return rob3(root, l, r);
    }

参考：

https://discuss.leetcode.com/topic/40847/simple-c-solution