?House Robber III

Posted 2020-07-24 Hygeia

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.

 

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 ];
    }
}