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.

class Solution {
    public int helper(TreeNode root, boolean take, HashMap<TreeNode, int[]> map) {
        if (root == null)
            return 0;
        if (map.containsKey(root)) {
            if (take)
                return Math.max(map.get(root)[0], map.get(root)[1]);
            return map.get(root)[0];
        }
        if (take) {
            int option1 = helper(root.left, true, map) + helper(root.right, true, map);
            int option2 = root.val + helper(root.left, false, map) + helper(root.right, false, map);
            int[] temp = new int[] { option1, option2 };
            map.put(root, temp);
            return Math.max(option2, option1);
        } else {
            return helper(root.left, true, map) + helper(root.right, true, map);
        }
    }

    public int rob(TreeNode root) {
        HashMap<TreeNode, int[]> map = new HashMap<>();
        return helper(root, true, map);
    }
}