Can the height of a binary search tree be less than that of a red-black tree?

This is a question from the book Algorithms by Robert Sedgewick and Kevin Wayne.

"Find a sequence of keys to insert into a BST and into a red-black BST such that the height of the BST is less than the height of the red-black BST, or prove that no such sequence is possible."

I think that in most cases, if not all, the height of a RBT is less than the height of a BST because the RBT ensures balance. However, is it possible for the height of a BST to be less than the height of a RBT? If so, how? Else, how to prove it?

The fact that a red-black tree is balanced just means that if it contains $n$ nodes then its depth is $O(\log n)$. There is no requirement for the depth to be optimal (i.e. $\lceil \log_2 (n+1) \rceil-1$), only for it to be optimal up to constants. Therefore it is definitely possible for a red-black tree to be deeper than some other binary search tree.

The advantage of balanced trees is that they are guaranteed to be balanced even after operations on the data structure. Since the running time of operations is usually linear in the depth, it is enough for the depth to be $O(\log n)$ rather than the optimal $\approx \log_2 n$.

StackExchange Computer Science Q#52814, answer score: 2