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.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int rob(TreeNode* root) {
        int use = 0;
	    int notuse = 0;

	    visit(root, use, notuse);
	    return max(use, notuse);
    }
private:
    void visit(TreeNode *root, int &use, int &notuse)
    {
	    if (root == NULL)
	    {
		    return;
	    }

	    int leftuse = 0;
	    int leftnotuse = 0;
	    int rightuse = 0;
	    int rightnotuse = 0;
	    visit(root->left, leftuse, leftnotuse);
	    visit(root->right, rightuse, rightnotuse);

	    use = root->val + leftnotuse + rightnotuse;
	    notuse = max(leftuse, leftnotuse) + max(rightuse, rightnotuse);
    }
};