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.

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     Map<TreeNode, Integer> map;
12     public int rob(TreeNode root) {
13         map = new HashMap<TreeNode, Integer>();
14         return helper(root);
15     }
16     
17     public int helper(TreeNode cur) {
18         if (cur == null) return 0;
19         if (map.containsKey(cur)) return map.get(cur);
20         
21         int val = 0; //initialize
22         if (cur.left != null) {
23             val += helper(cur.left.left) + helper(cur.left.right);
24         }
25         if (cur.right != null) {
26             val += helper(cur.right.left) + helper(cur.right.right);
27         }
28         int curOpt = Math.max(val+cur.val, helper(cur.left)+helper(cur.right));
29         map.put(cur, curOpt);
30         return curOpt;
31     }
32 }

 1 public int rob(TreeNode root) {
 2     int[] res = robSub(root);
 3     return Math.max(res[0], res[1]);
 4 }
 5 
 6 private int[] robSub(TreeNode root) {
 7     if (root == null) {
 8         return new int[2];
 9     }
10     
11     int[] left = robSub(root.left);
12     int[] right = robSub(root.right);
13     
14     int[] res = new int[2];
15     res[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
16     res[1] = root.val + left[0] + right[0];
17     
18     return res;
19 }