Is it always possible to turn one BST into another using at most O(n) tree rotations?

Question

This earlier question asks whether it's always possible to turn one BST for a set of values into another BST for the same set of values purely using tree rotations (the answer is yes). However, is it always possible to do this using at most O(n) total tree rotations?

Answer 1

Yes, it is always possible to turn one BST into another using at most O(n) tree rotations. This answer follows the same general approach as the other answer by picking some canonical tree shape T* and bounding the number of rotations needed to turn an arbitrary tree into our canonical tree. Then you can turn an arbitrary tree T₁ into another tree T₂ by transforming T₁ into T* and then transforming T* into T₂.

As suggested in comments, you can choose your canonical tree to be a degenerate linked list. For trees of n nodes, this upper bounds the number of rotations needed at 2n−2 .

In the paper Rotation Distance, Triangulation, and Hyperbolic Geometry , Daniel Sleator, Robert Tarjan, and William Thurston proved that the rotation distance between any two binary trees of n nodes is at most 2n−6 (better than the bound we get when transforming into a linked list).

At a high level, they did this by introducing a way to represent any binary tree as a polygon triangulation , where a tree rotation has a corresponding triangulation operation. Then, instead of reasoning about binary trees in their usual representation, the paper picks a canonical triangulation and shows how to transform an arbitrary triangulation into their desired one.

The canonical triangulation they chose is one where all diagonals emanate from a single vertex in a fan-like shape, which ends up corresponding to a somewhat unintuitive binary tree shape (a generalization of linked lists that also includes diamond shaped trees consisting of a root, a left child whose right child is a linked list, and a right child whose left child is a linked list).

It's a very cool technique that illustrates the power of isometries in data structures, showing how changing our representation can give us a new way of approaching a problem. Some friends and I recently put together a writeup walking through Sleator, Tarjan, and Thurston's proof if you would like to explore this in more detail.

Answer 2

Yes, this is always possible. I fear that the best I can do right now is give you a silly algorithm that proves it's possible, though I suspect that there must be a much better way to do this.

The Day-Stout-Warren algorithm is an algorithm that, starting with any BST, uses tree rotations to convert it to a perfectly balanced BST. It runs in time O(n) and does O(n) total rotations.

So suppose that you want to turn one tree T ₁ into another tree T ₂ using tree rotations. Run Day-Stout-Warren on both trees to convert them to the same balanced tree T*, and record the rotations that you needed to make in both cases. Then you can turn T ₁ into T ₂ by first running all the rotations needed to perfectly balanced T ₁ , then running the reverse of the rotations needed to turn T ₂ into a balanced tree. This turns T ₁ into T* and then turns T* into T ₂ . Since the Day-Stout-Warren algorithms makes only O(n) total rotations, this too makes only O(n) total rotations.

I feel like there has to be a better way to do this, but I'm not sure off the top of my head how to achieve this. If I think of anything, I'll let you know!