Suppose we are given a balanced binary search tree $T$, AVL tree, now
we want split $T$ at a arbitrary node. Can it be done in $O(log n)$?
First i read following links:
Split in AVL tree with complexity $O(log n)$
Trying to understand a way to split an AVL tree in O(log n)
Now i think, because of height of our tree $T$ is $O(log n)$ then after splitting at a arbitrary node, it is sufficient to re-balance our tree $T$ start from one of leaf node, in this manner we do rotation and because of each level want constant number of rotation to be balanced, after $O(log n)$ will be in root of $T$ and our tree $T$ is balanced. Are my argument is true?