How many non-isomorphic full binary trees are there?

Question

A full binary tree is a rooted tree where every node has either zero children or two children. My question is, how many non-isomorphic full binary trees are there, countably many or uncountably many?

Or more formally, what is the maximum cardinality a set can have if its elements are all full binary trees and any two elements are non-isomorphic?

Answer 1

There are $2^{\aleph_0}$ full binary trees up to isomorphism. This is obviously an upper bound, since every full binary tree is countable. To see that it can be achieved, given a binary tree $T$, consider the set $S\subseteq\mathbb{N}$ consisting of those $n$ such that every node in the $n$th level of $T$ has two children. It is easy to see that every subset $S\subseteq\mathbb{N}$ which contains $0$ can be realized by some full binary tree $T$, and so this gives $2^{\aleph_0}$ different isomorphism types of full binary trees.

Answer 2

Let $$T=\{0^n: n\in\mathbb{N}\}\cup\{0^n1: n\in\mathbb{N}\}.$$ $T$ consists of a single branch with a bunch of length-$1$ "spines" sticking out.

Now given $A\subseteq\mathbb{N}$, let $$T_A=T\cup\{0^n10: n\in A\}\cup\{0^n11: n\in A\}.$$ That is, we stick a "v" onto each "spine" corresponding to an element of $A$. Each $T_A$ is a full binary tree in your sense and we have $$T_A\cong T_B\iff A=B.$$ So there are continuum-many isomorphism types of full binary trees.