I am doing a quiz and here is one of the question asked:
You have implemented some recursive algorithm on a balanced binary search tree, which, in part, requires recursing on the left and then recursing on the right (similar to an in-order traversal). If there are N nodes in the tree, which of the following is true about your space complexity?
There are four answers to choose from.
A. It is exactly O(log N).
B. It is between O(1) and O(N).
C. It is at least O(log N), but could be worse.
D. It is at least O(N log N).
Given that we have a balanced binary search tree, I thought that the space complexity for the call stacks should be exactly O(log N). Why is the answer C instead?
YumekaMengjiaLYU your understanding of the space complexity for a recursive algorithm on a balanced binary search tree is mostly correct, but there’s an important nuance to consider. When you perform a traversal like in-order traversal, the maximum depth of the recursive calls corresponds to the height of the tree. For a balanced binary search tree, this height is indeed ( O(\log N) ). Therefore, the space complexity for the call stack during the recursion would typically be ( O(log N) ) in the best case.
However, the answer is C because while the space complexity is at least ( O(log N) ), it could potentially be worse in certain scenarios. For instance, if the tree were not perfectly balanced or if the algorithm were to use additional data structures that depend on the number of nodes, the space complexity could increase. Thus, the statement in option C accurately reflects that while the space complexity starts at ( O(log N) ), it could also scale up depending on the specific implementation details or the nature of the input data, making it a more comprehensive choice than A.