# Exercise 13.2.4

Show that any arbitrary $n$-node binary search tree can be transformed into any other arbitrary $n$-node binary search tree using $\O(n)$ rotations. (Hint: First show that at most $n-1$ right rotations suffice to transform the tree into a right-going chain.)

An informal argument:

Let's that the right-going chain of nodes in the tree, that is, nodes that can be reached by following only right-going edges. This right-going chain either contains the whole tree, or there are some nodes that are left children of the chain. If we perform a right rotation on such a child and its parent in the chain, the length of the chain increases by one and the number of nodes not in the chain gets reduced by one. We can keep iterating, until the tree is a right-going chain. There would be at most $n - 1$ nodes, so we can perform this in $\O(n)$ rotations.

It should be noted that this operation is reversible. That is, we can apply the symmetric left rotation in reverse order to obtain the original tree from the resulting right-going chain. Since there is only one possible right-going chain (sorting the elements in increasing order), we can transform any tree to another by first transforming it to a right-going chain, and then applying the reverse transformation that the target tree will need to become a right-going chain.