Rank of root in AVL tree

Question

Find a function that bounds from above and below (asymptotic) the rank of the root r in an AVL tree,
i.e. find a function $f(n)$ so there exists a constant $c>0$ that for every AVL tree with $n$ nodes so $c * f(n)\le rank(r)\le n-c*f(n) $.

$rank(r)$ gets his maximum value when his sub-left tree contains maximal number of nodes and because it's an AVL tree, for tree with height $h$ the sub left tree will be with height of $h-1$ and the number of nodes will be maximal when the sub tree is a full tree with $2^{h-2}-1$ nodes hence we get:
$rank(r)\leq 2^{h-2}-1+1=2^{h-2}$

On the other hand, $rank(r)$ gets his minimal value when his sub left tree contains a minimal number of nodes and it happens when the sub left tree is a fibonacci tree with height of $h-2$, therefore
$rank(r) \geq \phi ^{h-2}+1$.

Now I need to find f as a function of the number of nodes $n$ and not the height $h$. Am I in the right direction?
Would apricciate any help how to continue, Thanks.

Answer 1

First of all, it's not $2^{h-2}$, it's $2^h$.

The number of nodes $n$ in a full binary tree, is at most $n=2^{h+1}-1$, where $h$ is the height of the tree. A tree consisting of only a root node has a height of 0.

Straight outta wiki https://en.wikipedia.org/wiki/AVL_tree