Time complexity of insertion in binary search tree
It can't   be $O(n)$.
Suppose $n=2^k$,and for given $n$ elements, we inserted $\frac{n}{8}$ and $T$ is balanced. Now suppose you insert next $\frac{n}{8}$, But suppose when you inserted
$\frac{1}{2}\frac{n}{8}$ elements that  those make a chain with length $O(n)$. So for next $\frac{1}{2}\frac{n}{8}$    elements you maybe have cost $O(n)$ for each insertion. As a result the time complexity of $n$ insertion is $O(n^2)$.