Interview Question: Merge two sorted singly linked lists without creating new nodes

Node MergeLists(Node list1, Node list2) {
  if (list1 == null) return list2;
  if (list2 == null) return list1;

  if (list1.data < list2.data) {
    list1.next = MergeLists(list1.next, list2);
    return list1;
  } else {
    list2.next = MergeLists(list2.next, list1);
    return list2;
  }
}

Recursion should not be needed to avoid allocating a new node:

Node MergeLists(Node list1, Node list2) {
  if (list1 == null) return list2;
  if (list2 == null) return list1;

  Node head;
  if (list1.data < list2.data) {
    head = list1;
  } else {
    head = list2;
    list2 = list1;
    list1 = head;
  }
  while(list1.next != null) {
    if (list1.next.data > list2.data) {
      Node tmp = list1.next;
      list1.next = list2;
      list2 = tmp;
    }
    list1 = list1.next;
  } 
  list1.next = list2;
  return head;
}

Node MergeLists(Node node1, Node node2)
{
   if(node1 == null)
      return node2;
   else (node2 == null)
      return node1;

   Node head;
   if(node1.data < node2.data)
   {
      head = node1;
      node1 = node1.next;
   else
   {
      head = node2;
      node2 = node2.next;
   }

   Node current = head;
   while((node1 != null) ||( node2 != null))
   {
      if (node1 == null) {
         current.next = node2;
         return head;
      }
      else if (node2 == null) {
         current.next = node1;
         return head;
      }

      if (node1.data < node2.data)
      {
          current.next = node1;
          current = current.next;

          node1 = node1.next;
      }
      else
      {
          current.next = node2;
          current = current.next;

          node2 = node2.next;
      }
   }
   current.next = NULL // needed to complete the tail of the merged list
   return head;

}