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。