337. House Robber III java solutions

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.

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 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int rob(TreeNode root) {
12         int[] res = Robdfs(root);
13         return Math.max(res[0],res[1]);
14     }
15
16     public int[] Robdfs(TreeNode root){
17         if(root == null) return new int[2];
18         int[] left = Robdfs(root.left);
19         int[] right = Robdfs(root.right);
20         int[] ans = new int[2];
21         ans[0] = left[1] + root.val + right[1];
22         ans[1] = Math.max(left[0], left[1]) + Math.max(right[0],right[1]);
23         return ans;
24     }
25 }

DFS中，root为null时，返回长度为2的空数组；
建立结果数组res时，res[0]是包括根节点的情况，res[1]是不包含根节点的情况。而非按左右子树来进行划分的。