question about the construction of BSTs using a repeated sequence of rotations
Actually any BST can be transformed into balanced BST using $\mathcal{O}(n)$ rotations.

High level explanation: We convert given BST to right going chain(see fig-1 below). Then we convert that chain to balanced BST.

First convert given BST into right-going chain. Which can be accomplished using $n-1$ right rotations as follow.

Consider initial-right-chain $C$ of $root$ and all node that can be reachable from $root$ by following only right pointers.

Now, note that for any node $N$ in $C$ left child of node $N$ can be added into $C$ by right-rotating $N$. In doing that we don't loss any node from $C$.

So, by at most $n-1$ right rotation all node will be in $C$ because root will always be in $C$ from start.

Now, consider sequence of rotations $R: r_1, r_2, ..., r_m$ required to convert tree $T_1$ to right-going chain.

Now, consider sequence of rotations $S: s_1, s_2, ..., s_k$ required to convert balanced BST (containing same number of node as $T_1$) to right_going chain.

Then apply $r_1, r_2, ..., r_m, s_k', s_{k-1}', ..., s_1'$ to $T_1$. Here $s_i'$ denotes inverse operation of $s_i$.

Which will transform $T_1$ to balanced BST using at most $2n-2$ rotations.



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 figure - 1




Ref: CLRS 3e exercise 13.2.4