= Branch point of a tree
{title2=$\deg(v)\ge3$}
= Branch points of a tree
{synonym}
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>.<Graph Theory by Reinhard Diestel>[Third edition2005, section1.5, pp.13–15: trees and unique paths; full text https://webhomes.maths.ed.ac.uk/~v1ranick/papers/diestel1.pdf]{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 gives
$$
L=2+\sum_{\deg(v)\geq3}(\deg(v)-2),
$$
where $L$ counts degree-one <tree leaves>. Subtract two per vertex from $\sum_v\deg(v)=2(|V|-1)$: 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.
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