there are no edges joining the vertices of the same set $X_i$ and any two vertices from different $X_i,X_j$, $1\leqslant i,j \leqslant k$ are adjacent. In this case we write $G=G[X_1,...,X_k]$.

We shall say that $G$ is a [*complete multipartite graph*]{} if there exists $k$ such that $G$ is complete $k$-partite.

(root) \[solid node, label=right:[$x_0$]{}\] at (0,0) child [node(1) \[solid node, label=above:[$x_6$]{}\] ]{} child [node(2) \[solid node, label=right:[$x_5$]{}\] ]{} child [node(3) \[solid node, label=above:[$x_4$]{}\] ]{} child [node(3) \[solid node, label=above:[$x_3$]{}\] ]{} child [node(3) \[solid node, label=above:[$x_2$]{}\] ]{} child [node(3) \[solid node, label=above:[$x_1$]{}\] ]{};

In particular, a *star* is a complete bipartite graph $G[X_1, X_2]$ with $|X_1| = 1$ or $|X_2| = 1$ [@BM p. 4].

\[d14\] Let $(X,d)$ be a finite ultrametric space with $|X| \geqslant 2$. Define a graph $G_{D, X}$ as $V(G_{D, X}) = X$ and $$(\{u,v\}\in E(G_{D, X}))\Leftrightarrow (d(u,v)=\operatorname{diam}X)$$ for all $u$, $v \in V(G_{D, X})$. We call $G_{D, X}$ the *diametrical graph* of $X$.

\[t13\] Let $(X,d)$ be a finite ultrametric space, $|X|\geqslant 2$. Then $G_{D,X}$ is complete multipartite.

For every nonempty, finite ultrametric space $(X, d)$ we can associate a labeled rooted tree $T_X=T_X(r,l)$ with $r=X$ and $l\colon V(T_{X})\to \mathbb{R}^{+}$ by the following rule (see 