Question

I’m interested in the worst-case efficiency of stepping forwards and backwards through binary search trees.

Unbalanced tree:

   5
  /
 1 
  \
   2
    \
     3
      \
       4

It looks like the worst case would be 4->5, which takes 4 operations.

Balanced tree:

   2
  / \
 1   4
    / \ 
   3   5

Worst case is 2->3, which takes 2 operations.

Am I right in thinking that the worst case for any BST is O(height-1), which is O(log n) for balanced trees, and O(n-1) for unbalanced trees?

Answer 1

Am I right in thinking that the worst case for any BST is O(height-1), which is O(log n) for balanced trees, and O(n-1) for unbalanced trees?

Yes, you will only ever need to go up or down when travelling from k to k+1, never both (because the invariant is left child < parent < right child).

Although O(height-1) can be written O(height) (and similarly for O(n)).