I am working on a problem that requests to analyze the size of the intersection of two binary trees which are chosen independently and uniformly at random.
Formally speaking, consider an underlying complete binary tree $\text{CBT} = (V, E)$ of nodes $|V| = N = 2^L - 1$ (i.e., $L$ height in total) with root node $v_r \in V$. Given $n \leq N$, let \begin{equation} \mathcal{T}_{n} := \{T ~ \text{binary subtree in $\text{CBT}$} ~|~ T ~ \text{rooted at $v_r$},~ |T| = n\} \end{equation} be the set of subtrees of size $n$ in $\text{CBT}$ with root node $v_r$. Uniformly and independently pick two subtrees $T, T' \in \mathcal{T}_{n}$ at random. The target is to compute the probability of the size of the intersection of these two random binary trees $T, T'$, i.e., \begin{equation} \mathbb{P}_{T, T' \sim \text{Uni}[\mathcal{T}_n]} (|T \cap T'| = s), ~~~~ \text{for any $s \leq n$.} \end{equation}
My previous idea of computing this probability is by counting the number of binary trees $T$ which intersects with a given/fixed binary tree $T_*$ on $s$ nodes. One related paper is http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.722.412&rep=rep1&type=pdf which (eq. 2.38) provides an asymptotic expansion on the number of subtrees with nodes $k$ and height $h$. However, for different choices of $T_*$, the number of binary trees $T$ that satisfies $|T \cap T_*| = s$ is very different. The above paper is not sufficient to answer my question.