k successive calls to tree successor in bst

Question

Prove that K-successive calls to tree successor takes O(k+h) time. Since each node is visited atmost twice the maximum bound on number of nodes visited must be 2k. The time complexity must be O(k). I dont get where is the factor of O(h) coming. Is it because of nodes which are visited but are not the successor. I am not exactly able to explain myself how is the factor O(h) is involved in the whole process

PS:I know this question already exists but I was not able to understand the solution.

Answer 1

Plus in the O(k+h) notation is an alternative form of writing O(MAX(k, h)).

Finding a successor once could take up to O(h) time. To see why this is true, consider a situation when you are looking for a successor of the rightmost node of the left subtree of the root: its successor is at the bottom of the right subtree, so you must traverse the height of the tree twice. That's why you need to include h in the calculation: if k is small compared to h, then h would dominate the timing of the algorithm.

The point of the exercise is to prove that the time of calling the successor k times in a row is not O(k*h), as one could imagine after observing that a single call could take up to O(h). You prove it by showing that the cost of traversing the height of the tree is distributed among the k calls, as you did by noting that each node is visited at most twice.