1. 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.

This one has to be recursive, cause you don't really know which is going to be picked during the process. RobValue(Root)=Max of

1. RobValue(Root.left) + RobValue(Root.right)
2. RobValue(Root.left without left child) + RobValue(Root.right without right child) + Root.val;

The trick of this question is to return a pair of value from recursive call, res[2], first value is the max rob value from the current node, and res[1] is the max rob value without rob current node.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int rob(TreeNode root) {
        return rob_(root)[0];
    }

    private int[] rob_(TreeNode root){
        int[] res = {0 /*max of rob current sub tree*/, 
                    0 /* max of rob left child and right child*/};
        if(root != null){
            int[] left = rob_(root.left);
            int[] right = rob_(root.right);
            res[1] = left[0] + right[0];
            res[0] = Math.max(res[1], left[1] + right[1] + root.val);
        }

        return res;
    }
}