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 for adding this problem and creating all test cases.
树形dp
- rob_root = max(rob_L + rob_R , no_rob_L + no_nob_R + root.val)
- no_rob_root = rob_L + rob_R
-
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ //tree DPpublic class Solution { public int rob(TreeNode root) { return dfs(root)[0]; } public int[] dfs(TreeNode root){ if(root == null) return new int[2]; int[] left = dfs(root.left); int[] right = dfs(root.right); int dp[] = new int[2]; dp[1] = left[0] + right[0]; dp[0] = Math.max(dp[1] , left[1] + right[1] + root.val); return dp; }}