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:

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

Example 2:

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

题目含义：不能连续抢直接相连的两个节点。即例2中，抢了3就不能抢4，5。问最多能抢好多。

思路：给一个二叉树，求它不直接相连的节点的val和最大为多少。

如果抢了当前节点，那么它的左右孩子就肯定不能抢了。
如果没有抢当前节点，左右孩子抢不抢取决于左右孩子的孩子的val大小。

 1     private int[] dfs(TreeNode root)
 2     {
 3         int[] result = {0,0}; //result[0]表示抢当前节点 result[1]表示不抢当前节点
 4         if (root == null) return result;
 5         int[] leftResult = dfs(root.left);
 6         int[] rightResult = dfs(root.right);
 7         result[0] = leftResult[1] +rightResult[1] + root.val;//抢了当前节点，它的左右孩子就不可以抢了
 8         result[1] = Math.max(leftResult[0],leftResult[1]) + Math.max(rightResult[0],rightResult[1]);//不抢当前节点，左右孩子可抢可不抢
 9         return result;
10     }
11     
12     public int rob(TreeNode root) {
13         int[] result = dfs(root);
14         return Math.max(result[0],result[1]);        
15     }