Is it always possible to turn one BST into another using tree rotations?

Question

Given a set of values, it's possible for there to be many different possible binary search trees that can be formed from those values. For example, for the values 1, 2, and 3, there are five BSTs we can make from those values:

1      1      2      3      3
 \      \    / \    /      /
  2      3  1   3  1      2
   \    /           \    /
    3  2             2  1

Many data structures that are based on balanced binary search trees use tree rotations as a primitive for reshaping a BST without breaking the required binary search tree invariants. Tree rotations can be used to pull a node up above its parent, as shown here:

                rotate
         u      right           v
        / \     ----->         / \
       v   C                  A   u
      / \       <-----           / \
     A   B      rotate          B   C
                 left

Given a BST containing a set of values, is it always possible to convert that BST into any arbitrary other BST for the same set of values? For example, could we convert between any of the five BSTs above into any of the other BSTs just by using tree rotations?

Laevo

Answer 1

The answer to your question depends on whether you are allowed to have equal values in the BST that can appear different from one another. For example, if your BST stores key/value pairs, then it is not always possible to turn one BST for those key/value pairs into a different BST for the same key/value pairs.

The reason for this is that the inorder traversal of the nodes in a BST remains the same regardless of how many tree rotations are performed. As a result, it's not possible to convert from one BST to another if the inorder traversal of the nodes would come out differently. As a very simple case, suppose you have a BST holding two copies of the number 1, each of which is annotated with a different value (say, A or B). In that case, there is no way to turn these two trees into one another using tree rotations:

       1:a            1:b
         \             \
         1:b           1:a

You can check this by brute-forcing the (very small!) set of possible trees you can make with the rotations. However, it suffices to note that an inorder traversal of the first tree gives 1:a, 1:b and an inorder traversal of the second tree gives 1:b, 1:a. Consequently, no number of rotations will suffice to convert between the trees.

On the other hand, if all the values are different, then it is always possible to convert between two BSTs by applying the right number of tree rotations. I'll prove this using an inductive argument on the number of nodes.

As a simple base case, if there are no nodes in the tree, there is only one possible BST holding those nodes: the empty tree. Therefore, it's always possible to convert between two trees with zero nodes in them, since the start and end tree must always be the same.

For the inductive step, let's assume that for any two BSTs of 0, 1, 2, .., n nodes with the same values, that it's always possible to convert from one BST to another using rotations. We'll prove that given any two BSTs made from the same n + 1 values, it's always possible to convert the first tree to the second.

To do this, we'll start off by making a key observation. Given any node in a BST, it is always possible to apply tree rotations to pull that node up to the root of the tree. To do this, we can apply this algorithm:

while (target node is not the root) {
    if (node is a left child) {
        apply a right rotation to the node and its parent;
    } else {
        apply a left rotation to the node and its parent;
    }
}

The reason that this works is that every time a node is rotated with its parent, its height increases by one. As a result, after applying sufficiently many rotations of the above forms, we can get the root up to the top of the tree.

This now gives us a very straightforward recursive algorithm we can use to reshape any one BST into another BST using rotations. The idea is as follows. First, look at the root node of the second tree. Find that node in the first tree (this is pretty easy, since it's a BST!), then use the above algorithm to pull it up to the root of the tree. At this point, we have turned the first tree into a tree with the following properties:

1. The first tree's root node is the root node of the second tree.
2. The first tree's right subtree contains the same nodes as the second tree's right subtree, but possibly with a different shape.
3. The first tree's left subtree contains the same nodes as the second tree's left subtree, but possibly with a different shape.

Consequently, we could then recursively apply this same algorithm to make the left subtree have the same shape as the left subtree of the second tree and to make the right subtree have the same shape as the right subtree of the second tree. Since these left and right subtrees must have strictly no more than n nodes each, by our inductive hypothesis we know that it's always possible to do this, and so the algorithm will work as intended.

To summarize, the algorithm works as follows:

1. If the two trees are empty, we are done.
2. Find the root node of the second tree in the first tree.
3. Apply rotations to bring that node up to the root.
4. Recursively reshape the left subtree of the first tree to have the same shape as the left subtree of the second tree.
5. Recursively reshape the right subtree of the first tree to have the same shape as the right subtree of the second tree.

To analyze the runtime of this algorithm, note that applying steps 1 - 3 requires at most O(h) steps, where h is the height of the first tree. Every node will be brought up to the root of some subtree exactly once, so we do this a total of O(n) times. Since the height of an n-node tree is never greater than O(n), this means that the algorithm takes at most O(n²) time to complete. It's possible that it will do a lot better (for example, if the two trees already have the same shape, then this runs in time O(n)), but this gives a nice worst-case bound.

Hope this helps!

Laevo

Consider removing the "you"'isms. – Photocopy

You can do it in linear time ... see #19625825. – Kalinin

Answer 2

For binary search trees this can actually be done in O(n).

Any tree can be "straightened out", ie put into a form in which all nodes are either the root or a left child.

This form is unique (reading down from root gives the ordering of the elements)

A tree is straightened out as follows:

• For any right child, perform a left rotation about itself. This decreases the number of right children by 1, so the tree is straightened out in O(n) rotations.

If A can be straightened out into S in O(n) rotations, and B into S in O(n) rotations, then since rotations are reversible one can turn A -> S -> B in O(n) rotations.

Aleras

Wonderful answer! – Jeromejeromy