• ?House Robber III


    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 ];
        }
    }
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  • 原文地址:https://www.cnblogs.com/hygeia/p/5579323.html
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