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:

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

Output: 7 
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \ 
 1   3   1

Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
        Map<TreeNode, Integer> map =new HashMap<>();
        return helper(root, map);
    }
    
    public int helper(TreeNode node, Map<TreeNode, Integer> map) {
        if (node == null) {
            return 0;
        }
        if (map.containsKey(node)) {
            return map.get(node);
        }
        int val = 0;
        if (node.left != null) {
            val += helper(node.left.left, map) + helper(node.left.right, map);
        }
        if (node.right != null) {
            val += helper(node.right.right, map) + helper(node.right.left, map);
        }
        val = Math.max(val + node.val, helper(node.right, map) + helper(node.left, map));
        map.put(node, val);
        return val;
    }
}