A node in a binary tree is an only child if it has a parent node, but has no sibling (note that the root is not qualified as an only child). The “loneliness rate” (LR) of a binary tree T is defined as follows:
$$ LR = \frac{\text{number of $T$ nodes that are only children}}{\text{number of $T$ nodes}} $$
Show that for any non-empty AVL tree $T$, $LR(T) \leq \frac{1}{2}$.
I'm a little confused... I know that complete or perfect binary trees are AVL. I also know that a perfect binary tree of height $h$ contains $2^{h + 1}-1$ nodes.
But I cannot demonstrate the number of nodes that are only children. Will I have to demonstrate the maximum and minimum number of children?