Q:

Number of vertices with degree one

Suppose a tree where $m$ is number of vertices with degree one in the tree . Prove if graph $G$ has less vertices with degree one than $m$ then $G$ is a cyclic graph and if graph $G$ has more vertices with degree one than $m$ then $G$ is a disconnected graph . 
Example : $4 , 3 , 2, 2 , 2 ,2 , 1 ,1,1,1,1$ is degree sequence of a tree . Therefore $4 , 3 , 2, 2 , 2 ,2 , 1 ,1,1$ belongs to a cyclic graph and $4 , 3 , 2, 2 , 2 ,2 , 1 ,1,1,1,1, 1 , 1$ belongs to an disconnected graph . 
I really have no idea for the proof .

A:

I must admit that I am not sure I got your question right, I'll give it a try anyway.
Notice that if $(G,E)$ is a tree, then $|E|+1=|G|$, and notice that $|E|$ is just the sum of the degree of every node divided by $2$.
Let $K$ be the sequence of the degrees of the nodes of a tree, like in your example above. If a sequence $K'$ of a graph $(G',E')$ only differs from $K'$ by the number of $1$'s, then $|E'|+1<|G'|$, since every new vertex only contributes $1/2$ to the number of edges. So the graph is disconnected (it is not difficult to show this: just proceed by contradiction).
If otherwise the sequence $K'$ has fewer $1$'s than $K$, the opposite is true, that is $|E'|+1>|G'|$. If the graph $(G',E')$ has $m$ connected components $(G'_i,E'_i)$, there is at least one of them having too many edges to be a tree, since $\sum|G_i|=|G'|$ and $\sum|E'_i|=|E'|$. Since we are assuming that this component is connected, it has to contain a cycle.

