Branch point of a tree ($\mathrm{deg}(v)\ge 3$)

A branch point of a tree is a vertex of vertex degree at least three in the underlying undirected tree. Removing it separates the remaining tree into at least three connected components of a graph. This is a structural property of the graph, independent of which root of a tree is chosen.

The component count has a short proof. If $v$ has $d$ neighbors, each neighbor belongs to a component of $T-v$. Two different neighbors cannot remain connected there, because that connecting graph path together with their two edges to $v$ would form a graph cycle. Every remaining vertex reaches one of the neighbors along its original route to $v$. Hence $T-v$ has exactly $d$ components. The proof uses the unique-path property of a tree.^[ref]

For an original example, take $E=\{ra,ab,ac\}$. Vertex$a$ has degree three, and deleting it leaves the three singleton components $r,b,c$. Rooting at$r$ gives$a$ two children. Rooting at$a$ gives the root three children. Its undirected degree remains three in both cases.

The distinction matters at a root with only two children: its degree is two, so it does not meet this article's branch-point convention, even though a rooted drawing visibly splits there.

For a finite tree with at least two vertices, the degree-sum formula also giveswhere $L$ counts degree-one tree leaves. Subtract two per vertex from ${\sum }_v\mathrm{deg}(v)=2(|V|-1)$: degree-one vertices contribute−1, degree-two vertices contribute zero, and branch points contribute their excess above two. A degree-four hub therefore gives $L=4$. A crossing in a drawing contributes nothing unless it is actually a graph vertex.