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

题目：前一题的继续，如果给定的是任意的二叉树，前题的代码还能用吗？

思路：显然，前一题不能用了。只好用层序遍历。

	public void connect(TreeLinkNode root) {
		if(root == null)
			return;
		TreeLinkNode parent = root,pre,next;
		while(parent != null){
			pre = null;
			next = null;
			while(parent != null){
				if(next == null)
					next = (parent.left == null) ? parent.right : parent.left;
				if(parent.left != null){
					if(pre != null){
						pre.next = parent.left;
						pre = pre.next;
					}else
						pre = parent.left;
				}
				if(parent.right != null){
					if(pre != null){
						pre.next = parent.right;
						pre = pre.next;
					}else
						pre = parent.right;
				}
				parent = parent.next;
			}
			parent = next;
		}
	}