@Arjun Sir, it would be really helpful to add the fact that :
Let $n_l$ and $n_r$ nodes be present in the left and right sub-trees respectively.
We have, $\frac{n_r}{2}\le n_l \le 2n_r$. Without loss of generality, let the left sub-tree have greater number of nodes $(2n_r\ nodes)$. Then, $n_r + 2n_r + 1 = n$. Thus we get, $n_l = {2(n-1) \over 3}$ and $n_r = {n-1 \over 3}$.