Height of balanced binary search tree = logn
So there are logn levels in tree
At one node we need to find left and right sub tree leaves
The recurrence relation will be T(n) = left sub tree + right sub tree + root
T(n) = T(n/2)+T(n/2)+1 = 2T(n/2) + 1 = O(n)
There are logn levels so T(n) = O(nlogn)