Populating Next Right Pointers in Each Node II

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

Tips:

Level order traversal, 见注释

Code:

public class Solution {
    public void connect(TreeLinkNode root) {
        // Logic: 
        //        node moves to all nodes in tree in level order fashion
        //        needle keeps sewing together next level's children

        // levelHead is sentinel, keeps track of start node of next level
        TreeLinkNode node = root, levelHead = new TreeLinkNode(0);

        while(node != null){ // this loop is for different levels

            // needle is sewing next fields in current level
            // first time in a level, it is same as leavelHead (with null next)
            // but as soon as we get a non null child from node
            // needle threads that child into its next, 
            //      --------->>  thus making that child as next levelHead
            //
            TreeLinkNode needle = levelHead;

            // this loop moves node in current level
            while(node != null){

                if(node.left != null){
                    needle.next = node.left;
                    needle = needle.next;
                }
                if(node.right != null){
                    needle.next = node.right;
                    needle = needle.next;
                }

                node = node.next; // move node to next in same level, end up null at rightmost
            }
            // current level ended in node being null
            // take node fom sentinel's next, which is next levels start
            node = levelHead.next;

            // levelhead.next grabbed into node above, it has been used
            // so make it null so next time we dont grab again, 
            //    if all next lvl is null, it remains null prompting end of run
            levelHead.next = null;  
        }
    }
}