# 18.2 Big O vs. Worst Case

Consider the following statements about BSTs. Which of the following are true?

1. The worst-case height of a BST is $$\Theta(N)$$.
2. BST height is $$O(N)$$.
3. BST height is $$O(N^2)$$.

The answer is that all three statements are true. BSTs always have a height that is linear or better, and a linear height is obviously "less than" the quadratic upper bound in the last point.

However, a more tricky question is which of the three statements is *the most informative*.

The answer here is the first statement: it gives an *exact* upper and lower bound unlike the other statements. $$O(N)$$ could mean linear, logarithmic, square-root, or constant, but $$\Theta(N)$$ can only mean linear.

For an analogy, consider the following statements about the worst-case cost of a hotel room:

1. The most expensive room is $639/night.
2. The most expensive room is less than or equal to $2000/night.

Here, we see that the first statement gives us exact information, whereas the second statement does not. In the second statement, the most expensive room could be $2000, $10, or anywhere in between.

However, *both are statements about the worst case*. Applying this to asymptotic notation, this means that we can refer to the worst case with $$\Theta$$, $$O$$, or even $$\Omega$$. **Big O is not the same as the worst case!**

## Using Big O

If $$\Theta$$ is always more informative than $$O$$, then why do we bother using Big O notation at all? There are several reasons:

* We can make broader statements. For example, saying "binary search is $$O(\log N)$$ is correct, but saying "binary search tree is $$\Theta(log N)$$" would not be correct, since it can be constant in certain scenarios.
* Sometimes, it is not possible or extremely difficult to determine the exact runtime. In such cases, we would still like to provide a generalized upper bound.
