I'm afraid the author is writing for an audience that's familiar with generating functions. It's using a common trick in analytic combinatorics, the notation is

I'm afraid the author is writing for an audience that's familiar with generating functions. It's using a common trick in analytic combinatorics, the notation is actually "shorthand". Very informally, let T(z) be the generating function where the coefficient of the z^n term is the number of trees we are searching with exactly n nodes. We know from the functional description that such a tree can be: - a leaf; - a node with 1 child; - a node with 2 children; - a node with 3 children. A leaf is a tree with one node and that's it, so it "contributes" as a single tree of size 1. How many trees of size n are there that fall in the second case? Exactly as many as the total trees of size (n-1), hence this case "contributes" as zT(z). How many trees of size n fall in the third case? That's a bit harder to see, because now you have (n-1) nodes in total, split between the two branches. So you can have 1 in the left branch and n-2 in the right branch, 2 left n-3 right and so on. With a bit of basic combinatorics you can notice that this is exactly the convolution product of the series with itself, in analytical terms T^2(z). Hence the third case "contributes" for zT^2(z). The fourth case is similar. At this point we have constructed the inverse of the generating function we are looking for, i.e. z as a function of T(z) rather than the reverse, but luckily there's a standard trick for this: Lagrange inversion.