LeetCode_OJ【337】 House Robber III

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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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

做这个题目的时候参考了如下帖子：http://blog.csdn.net/tstsugeg/article/details/50867315

对于任意一个节点，有两种状态：偷或者不偷，我们用money[2]来表示这两种状态下以该节点为根节点的子树所能偷取到的最大价值，其中money[0]表示不偷该节点，money[1]表示偷该节点。同样的我们用lmoney[2],rmoney[2]这两个数组来表示该节点左右子树能偷得的最大价值，容易得出：

money[1] = lmoney[0] + rmoeny[0] + node.val; (不能偷相邻节点，所以如果本节点被偷，左右子节点都不能偷)；

money[0] = max(lmoney[0],lmoney[1]) + max(rmoney[0],rmoney[1]); (不偷本节点，左右子节点都可以偷，取左右节点两种情况最大值)；

具体的Java实现如下：

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

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