Here is what I need to prove:
Let $d_1,d_2,...,d_n$ be a sequence of natural numbers (>0). Show that $d_i$ is a degree sequence of some tree if and only if $\sum d_i = 2(n-1)$.
I know that:
1. for any graph $\sum_{v \in V}\ deg(v) = 2e$;
2. for any tree $e=v-1$.
From 1 and 2 it follows that for any tree $\sum_{v \in V}\ deg(v) = 2(v-1)$.
If I understand it correctly, this is only a half of the proof ($\rightarrow$), isn't it?
Any hints on how to prove it the other way?
Edit (induction attempt):
$n=1$, we have $d_1 = 2(1-1) = 0$ and $d_1$ is a degree sequence of a tree.
Let's assume the theorem holds for all $k
Is this proof correct or am I missing the point?