Binary search vs binary search tree

Question

What is the benefit of a binary search tree over a sorted array with binary search? Just with mathematical analysis I do not see a difference, so I assume there must be a difference in the low-level implementation overhead. Analysis of average case run time is shown below.

Sorted array with binary search
search: O(log(n))
insertion: O(log(n)) (we run binary search to find where to insert the element)
deletion: O(log(n)) (we run binary search to find the element to delete)

Binary search tree
search: O(log(n))
insertion: O(log(n))
deletion: O(log(n))

Binary search trees have a worst case of O(n) for operations listed above (if tree is not balanced), so this seems like it would actually be worse than sorted array with binary search.

Also, I am not assuming that we have to sort the array beforehand (which would cost O(nlog(n)), we would insert elements one by one into the array, just as we would do for the binary tree. The only benefit of BST I can see is that it supports other types of traversals like inorder, preorder, postorder.

Answer 1

Your analysis is wrong, both insertion and deletion is O(n) for a sorted array, because you have to physically move the data to make space for the insertion or compress it to cover up the deleted item.

Oh and the worst case for completely unbalanced binary search trees is O(n), not O(logn).

Answer 2

There's not much of a benefit in querying either one.

But constructing a sorted tree is a lot faster than constructing a sorted array, when you're adding elements one at a time. So there's no point in converting it to an array when you're done.