Average number of full nodes in rooted m-ary tree
Every $m$-ary tree is a node together with up to $m$ children, which are also $m$-ary trees. Let $T(x,y)$ be the generating function in which the coefficient of $x^ny^k$ is the number of $m$-ary trees with $n$ nodes and $k$ full nodes. Then
$$
T(x,y) = x(1 + T(x,y) + \cdots + T(x,y)^{m-1} + yT(x,y)^m).
$$
The generating function of the total number of $m$-ary trees is $A(x) = T(x,1)$, and the generating function of the total number of full nodes in $m$-ary trees is $B(x) = \frac{\partial}{\partial y} T(x,y)|_{y=1}$.

The generating function $A(x)$ satisfies the equation
$$
A(x) = x(1 + A(x) + \cdots + A(x)^m).
$$
The generating function $B(x)$ satisfies the equation
$$
B(x) = x(B(x) + 2A(x)B(x) + 3A(x)^2B(x) + \cdots + mA(x)^{m-1}B(x) + A(x)^m),
$$
and so
$$
B(x) = \frac{xA(x)^m}{1-x(1+2A(x)+3A(x)^2+\cdots+mA(x)^{m-1})}.
$$

You can try solving these equations.

For example, in the trivial case $m=1$, we have $A(x) = x(1 + A(x))$, and so
$$ A(x) = \frac{x}{1-x} = \sum_{n \geq 1} x^n. $$
It follows that
$$
B(x) = \frac{x\cdot \frac{x}{1-x}}{1-x} = \frac{x^2}{(1-x)^2} = \sum_{n \geq 1} (n-1)x^n.
$$
Therefore the average number of full nodes in $1$-ary trees with $n$ nodes is $\frac{n-1}{1} = n-1$.