337. 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.

Solution:
Similar with all previous House Robber problem, each node may have two states, robbed, not robbed,
Constrains:
If current node is robbed, we can't rob its children. 
If current node is not robbed, we can either rob or not rob its children depends on which option gives us the maximum profit. 
Recursion to get the maximum money.

 public int rob(TreeNode root) {  
     int[] res=getMoney(root);  
     return Math.max(res[0],res[1]);  
   }  
   public int[] getMoney(TreeNode root){  
     if(root==null) return new int[2];  
     int[] left=getMoney(root.left);  
     int[] right=getMoney(root.right);  
     int[] res=new int[2];  
     res[0]=root.val+left[1]+right[1];  
     res[1]=Math.max(left[0],left[1])+Math.max(right[0],right[1]);  
     return res;  
   }