Need help explaining answers The question is this: What is the size of the largest and smallest numbers of vertices of degree 1 (AKA Leaves) possible in an n-vertex tree, for n > 2?
The answer is of course
largest number: n-1
smallest number is: 2
I know this because if you make a tree with 3 vertices. 1 parent vertex with 2 child vertices then the number of leaves is 2 which is n-1 and this expands of course to trees with n-amount of vertices where n>2.
Likewise the smallest number I calculated using the above. If you make a tree with 3 vertices. The number of leaves is 2. This is the smallest tree and thus the smallest number is 2 as any other tree with >3 vertices would have more leaves then 2.
However, I feel like these aren't very good explanations. Are they? I just feel like I'm saying its n-1 and 2 just cause that's how it is which is a terrible argument. If someone could help me explain in better mathematical terms why the answers are n-1 and 2, it would greatly appreciated.
 A: Your argument is completely valid, it's just need a bit of "polishing".
Let's look at the first question,

What is the size of the largest numbers of vertices of degree 1 (AKA Leaves) possible in an $n$-vertex tree, for $n > 2$?

Call $a$ this number. Then by definition we must have $a\in \{0,\ldots,n\}$. It's pretty simple to argue that you cannot have $a=n$. For a very precise argument you could say:
Suppose that there exist a tree $T$ on $n$ vertices with $n$ leaves, i.e. all vertices have degree $1$. Let $m$ the number of edges of $T$, $\{v_1,\ldots,v_n\}$ its vertices, and $deg(v)$ the degree of $v$. Then by hand-shaking lemma,
$$ \sum_{i=1}^n deg(v_i)=2m.$$
Therefore $m=n/2<n-1$ for $n>2$. A contradiction with $T$ being a tree (any tree on $n$ vertices has exactly $m=n-1$ edges). Therefore we must have $a\leq n-1$.
Then you exhibit an explicit example : the start graph $K_{1,n-1}$, made of $1$ central vertex, and $n-1$ leaves. Therefore $a\geq n-1$. And you conclude $a=n-1$.
The other problem (smallest number of leaves) follows the same logic : you exhibit the path $P_n$ with $2$ leaves, and if a tree $T$ has only $1$ leaf, then again by hand-shaking lemma
$$ 2m=1+\sum_{i=2}^n deg(v_i)\geq 1+2(n-1) > 2(n-1),$$
a contradiction with $m=n-1$.
