leetcode 337. House Robber III DP动态规划 + DFS深度有限遍历

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.

这道题意很简单，就是父亲节点和自己不可以同时抢劫，这里设置一个返回数字res，其中res[0]表示抢劫当前节点， res[1]不抢劫当前节点，直接DFS递归遍历即可实现。

建议和这道题leetcode 198. House Robber 入室抢劫 + DP求解 和 这道题 leetcode 213. House Robber II 入室抢劫 抢劫问题 一起学习。

代码如下：

/*class TreeNode {      int val;      TreeNode left;      TreeNode right;      TreeNode(int x) { val = x; }}*//* * 这其实也是DP的解决方法之一， * */class Solution{    public int rob(TreeNode root)    {        if(root==null)            return 0;        int []res=getRes(root);        return Math.max(res[0], res[1]);    }    /*     * res[0]表示抢劫当前节点的记录的最大值     * res[1]表示不抢劫当前的结点的记录的最大值     * */    public int[] getRes(TreeNode root)     {        int[] res=new int[2];        if(root==null)            return res;        else         {            int[] left=getRes(root.left);            int[] right=getRes(root.right);            //抢劫当前节点，然后不抢劫其子节点            res[0]=root.val + left[1] + right[1];            //不抢劫当前节点，然后可以抢劫或者不抢劫其子节点            res[1]=Math.max(left[0], left[1]) + Math.max(right[0], right[1]);            return res;        }    }}