Below is a question I was asked in an Interview

What is the best case time complexity to find the height of a Binary Search Tree?

I answered it explaining using the below algorithm


if $v$ is a leaf, return 0

otherwise, return $\max(\mathrm{height}(v.\mathit{left}), \mathrm{height}(v.\mathit{right})) + 1$

So, in best case, my recurrence would become

$$T(n)=2T\left(\frac{n}{2}\right) + c.$$

Here $T(\frac{n}{2})$ is for each of the recursive calls, and $c$ for all the rest. So even best case complexity is $O(n)$.

Now, in the worst case, my recurrence would become


and this would be a case of a skewed BST. Still, here complexity remains $O(n)$.

So, in all cases, the time complexity to find the height of a BST remains $O(n)$.

Is my claim correct?

  • $\begingroup$ The skewed case is the worst case scenario, and this will be enough for calculating the Big-Oh of the problem. So $O(n)$ is indeed the correct claim. $\endgroup$
    – Sagnik
    Jul 30, 2018 at 5:57
  • $\begingroup$ @Sagnik-And the best case would also be $\Omega(n)$, right? $\endgroup$ Jul 30, 2018 at 7:15

1 Answer 1


Your algorithm runs in linear time on all inputs. The algorithm visits each node of the tree exactly once, and does $O(1)$ work per node. Therefore it runs in time $\Theta(n)$, where $n$ is the number of nodes.

The argument above is better than using recurrences, since it is more immediate. It also shows that if you had an arbitrary tree (not necessarily binary), the algorithm would still run in linear time.

  • $\begingroup$ More of an aggregate analysis...am i correct @YuvalFilums ? $\endgroup$ Jul 30, 2018 at 8:06
  • $\begingroup$ I'm not familiar with this term. $\endgroup$ Jul 30, 2018 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.