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 1Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3 / \ 4 5 / \ \ 1 3 1Maximum amount of money the thief can rob = 4 + 5 = 9.
Why this solution does not work?
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public int rob(TreeNode root) { List<Integer> res = new LinkedList<Integer>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); if(root==null) return 0; queue.add(root); while(!queue.isEmpty()) { int levelnum = queue.size(); //这层有几个TreeNode int levelval = 0; for(int i=0; i<levelnum;i++) { TreeNode node = queue.poll(); if(node.left!=null) queue.add(node.left); if(node.right!=null) queue.add(node.right); levelval = levelval + node.val; } res.add(levelval); } if(res.size()<=1) return res.size()==0?0:res.get(0); int[] dp = new int[res.size()]; //init dp[0] = res.get(0); //the second is should to be calculate dp[1] = res.get(0)>res.get(1)?res.get(0):res.get(1); for(int i=2;i<res.size();i++) dp[i] = Math.max(dp[i-1], dp[i-2]+res.get(i)); return dp[res.size() -1 ]; } }
转载于:https://www.cnblogs.com/hygeia/p/5579323.html
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