A balanced binary tree is a full binary tree in which every leaf is either at level l or l-1 for some positive integer l. The set of balanced binary trees is defined recursively by: Basis step: A single vertex is a balanced binary tree. The tree with two vertices, namely a root and a left child (a leaf) is a balanced binary tree. The tree with two vertices, namely a root and a right child (a leaf) is a balanced binary tree. Recursive step: A balanced binary tree T= T1 T2 consists of a new root r together with edges connecting the r to each of the roots, r1 and r2, of two balanced binary trees, T1 (the left subtree) and T2 (the right subtree), respectively, where the leaves of T1 and T2 are at either level l or l‐1 for some positive integer l.
Draw the balanced binary trees formed by 0, 1 and 2 applications of the recursive step of the recursive definition above.
Let $S,L,R$ be the BBTs with a single vertex, a root and a left child, and a root and a right child. These are the trees formed by zero applications of the recursive step. For trees formed by one application of the recursive step, we have a-priori nine possibilities: $$S\square S,S\square L,S\square R,L\square S,L\square L,L\square R,R\square S,R\square L,R\square R.$$ However, we need to check that the condition that all leaves be on two adjacent levels holds. Since the leaves of $S,L,R$ are all on levels $0,1$, this condition always holds. When considering trees formed by two applications, there will be cases in which the condition doesn't hold, for example $S\square(S \square L)$, in which leaves are at levels $1,2,3$.
The most difficult case, of two applications, I leave to you. There are two (symmetric) possibilities two consider: $\alpha\square(\beta\square\gamma)$ and $(\alpha\square\beta)\square\gamma$, where $\alpha,\beta,\gamma \in \{S,L,R\}$. Hence a-priori there are 54 possibilities, but as mentioned above, some of them will be ruled out.