Condition on degrees for existence of a tree 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 $k<n$. So we know that $d_1 + d_2 + ... + d_{n-1} = 2(n-2) $ and that it's a degree sequence of a tree. In order to add one vertex to this tree, so that it remains a tree we only can add a vertex of degree one (we connect it to any existing vertex with a single edge). By doing this our new sequence is $1,d_1,...,d_j+1,...,d_{n-1}$, where $j$ is the vertex to which we attached the new vertex (it's degree increments). Clearly this sums to $2(n-1)$.
Is this proof correct or am I missing the point?
 A: Hints for the other way round: 


*

*Do you know about mathematical induction?

*Can every $d_i$ be $\gt 1$?

A: From your fact (1) and the hypothesis $\sum d_i = 2(n-1)$, you know $e = n - 1$.  You need now only prove that any connected graph with $n - 1$ edges is a tree.
EDIT - Since you are going down an alternate path, I have a new suggestion.
Order the $d_i$ so that $d_1 \leq d_2 \leq \ldots \leq d_n$.  Construct a tree with degree sequence $d_2, \ldots, d_n$. You know $\sum_{i \neq 1} d_i = 2(n - 2)$ (why?).  Now try to construct your tree with degree sequence $d_1, \ldots, d_n$.  You have to be careful about how you add in your vertex with degree $d_1$ so that it remains a tree, but this can be done.
EDIT - The above is incorrect.  The idea remains to add in a vertex of degree $1$ to a tree with a specific degree sequence related to $d_1, \ldots, d_n$. But $\sum_{i \neq 1} d_i = 2n - 3$, so we cannot claim $d_2, \ldots, d_n$ is a degree sequence of a tree by the inductive hypothesis.  Of course, when we add in the vertex of degree $1$, we alter the degree of another vertex, right?
