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 '18 at 5:57
  • $\begingroup$ @Sagnik-And the best case would also be $\Omega(n)$, right? $\endgroup$ – user3767495 Jul 30 '18 at 7:15

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 ? $\endgroup$ – Navjot Singh Jul 30 '18 at 8:06
  • $\begingroup$ I'm not familiar with this term. $\endgroup$ – Yuval Filmus Jul 30 '18 at 8:27

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