Any tree degree sequence is a caterpillar degree sequence

Question

In this question Joffan comments

My intuition says that any degree sequence that corresponds to a tree can be laid out... in a caterpillar. That would be cool to have a proof of.

So I am asking and answering the question.

Answer 1

Theorem. Given any tree, there is a caterpillar with the same degree sequence.

Proof. Suppose the degree sequence consists of $d_1,d_2,\ldots,d_s$, all greater than or equal to $2$, and $e$ elements equal to $1$. This is the degree sequence of a tree so we have $$\langle\hbox{total degree}\rangle =2\langle\hbox{number of edges}\rangle =2\langle\hbox{number of vertices}\rangle-2\ ,$$ which gives $$d_1+\cdots+d_s+e=2s+2e-2$$ and hence $$e=d_1+\cdots+d_s-2s+2\ .$$ Now it is clear that the vertices of degree $d_1,\ldots,d_s$ can be formed into a chain with $s-1$ edges, and so the number of "available" edges is $$d_1+\cdots+d_s-2(s-1)\ ,$$ which is exactly $e$. So all the vertices of degree $1$ can be fitted around this chain, forming a caterpillar.