Populating Next Right Pointers in Each Node I or 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

나 옛날에 왜 그렇게 복잡했지?
ref http://www.cnblogs.com/springfor/p/3889327.html
관건은 효과적인 다음 P를 찾는 것이다

 public void connect(TreeLinkNode root) {  

        if (root == null) 

            return;  

  

        TreeLinkNode p = root.next;  

        while(p!=null){

            if(p.left!=null){

                p = p.left;

                break;

            }

            if(p.right!=null){

                p = p.right;

                break;

            }

            p = p.next;

        }

        if(root.right!=null)

            root.right.next = p;

        if(root.left!=null){

            if(root.right!=null){

                root.left.next = root.right;

            }else

                root.left.next =p;

        }

  

        connect(root.right);

        connect(root.left);

    }