?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 ];
}
}