Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /        2    3
     / \        4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /        2 -> 3 -> NULL
     / \        4-> 5 -> 7 -> NULL

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
        int level = 1;//记录需要出队的元素数，
        if(root == null ){
            return;
        }
        queue.offer(root);
        int count = 0;
        TreeLinkNode p = null;
        while(queue.size() > 0){
            int level1 = 0;//下一层需要出队的个数
            while(count < level - 1){
                p = queue.poll();
                p.next = queue.peek();
                if(p.left != null){
                    queue.offer(p.left);
                    level1++;
                }
                if(p.right != null){
                    level1++;
                    queue.offer(p.right);
                }
                count++;
            }
            p = queue.poll();
            if(p.left != null){
                queue.offer(p.left);
                level1++;
            }
            if(p.right != null){
                level1++;
                queue.offer(p.right);
            }
            level = level1;
            count = 0;
        }
    }
}

Populating Next Right Pointers in Each Node II