leetcode [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:

Input: [3,2,3,null,3,null,1]

     3
    / \
   2   3
    \   \ 
     3   1

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

Example 2:

Input: [3,4,5,1,3,null,1]

     3
    / \
   4   5
  / \   \ 
 1   3   1

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

题目大意：

抢劫房子可以获得的最大钱数，不能抢劫相邻的房子，房子的结构形成了一个二叉树结构。

解法：

采用动态规划的方法，每一个节点都有一个数组nums，nums[0]代表是抢这个节点的最大获利，nums[1]代表的则是不抢这个节点的最大获利。

java:

class Solution {
    private int[] dfs(TreeNode x){
        if (x==null) return new int[2];
        int[] left=dfs(x.left);
        int[] right=dfs(x.right);
        int[] res=new int[2];
        res[0]=left[1]+right[1]+x.val;
        res[1]=Math.max(left[0],left[1])+Math.max(right[0],right[1]);
        return res;
    }

    public int rob(TreeNode root) {
        int[] num=dfs(root);
        return Math.max(num[0],num[1]);
    }
}