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The idea is to have to trees that are linked by an edge "a", it is then removed and you relabel the two subtrees. It does not have to be exactly like that. The subtrees can be different. What matters it the general idea. Thanks! enter image description here

1 Answer 1

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Can this be done? Yes. Am I now being very creative in designing four different trees? No.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{fit}
\begin{document}
\begin{tabular}{cp{2cm}c}
\begin{tikzpicture}[remember picture,grow=west]
\node[circle,fill] (TL) {}
    child{ node[circle, fill = black] {}}
    child{ node[circle, fill = black] {}
    };
\node[circle,draw,dashed,fit=(TL)(TL-1)(TL-2)]{};   
\end{tikzpicture}       
& & 
\begin{tikzpicture}[remember picture,grow=east]
\node[circle,fill] (TR) {}
    child{ node[circle, fill = black] {}}
    child{ node[circle, fill = black] {}
    };
\node[circle,draw,dashed,fit=(TR)(TR-1)(TR-2)]{};   
\end{tikzpicture}       \\
\begin{tikzpicture}[remember picture,grow=west]
\node[circle,fill] (BL) {}
    child{ node[circle, fill = black] {}}
    child{ node[circle, fill = black] {}
    };
\node[circle,draw,dashed,fit=(BL)(BL-1)(BL-2)]{};   
\end{tikzpicture}       
& & 
\begin{tikzpicture}[remember picture,grow=east]
\node[circle,fill] (BR) {}
    child{ node[circle, fill = black] {}}
    child{ node[circle, fill = black] {}
    };
\node[circle,draw,dashed,fit=(BR)(BR-1)(BR-2)]{};   
\end{tikzpicture}       \\
\end{tabular}
\begin{tikzpicture}[overlay,remember picture]
\draw (TL) -- (TR) node[midway,above] {$a$};
\draw[dotted] (BL) -- (BR);
\end{tikzpicture}
\end{document}

enter image description here

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