LeetCode-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.

Analysis:

Two states at each root: rob root, not rob root.

Solution:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public class RobResult {
        int robProfit;
        int noRobProfit;
        public RobResult(int rp, int nrp){
            robProfit = rp;
            noRobProfit = nrp;
        }
    }
    
    public int rob(TreeNode root) {
        RobResult res = robRecur(root);
        return Math.max(res.robProfit,res.noRobProfit);        
    }

    public RobResult robRecur(TreeNode cur){
        if (cur==null){
            return new RobResult(0,0);
        }

        RobResult leftRes = robRecur(cur.left);
        RobResult rightRes = robRecur(cur.right);

        RobResult curRes = new RobResult(leftRes.noRobProfit+rightRes.noRobProfit+cur.val, Math.max(leftRes.noRobProfit,leftRes.robProfit)+Math.max(rightRes.noRobProfit,rightRes.robProfit));
        return curRes;
    }
}