It is known that a tree $T=(V,E)$ has at least $\Delta$ leaves, where $\Delta$ is the maximum degree of $T$. But how many vertices of specific degree at least $k$ can a tree have at most? I'm particularly interested in $k=3$, but also in general. Is there an easy way to see or count this?
Let $s$ be the number of vertices of degree 3 or more in any tree. I think it's easy to see that $s < |V|$, but this is probably not tight.
If we denote the number of vertices as $n$, from the hand shaking lemma and the fact that the number of edges in a tree is $n-1$, we have $$\sum_{v \in V} d_n = 2 \vert E \vert = 2(n-1)$$ If $s_k$ vertices have degree at-least $k$, then we have $$2(n-1) = \sum_{v \in V} d_n \geq s_k \times k +(n-s_k)$$ Hence, we get that $$s_k(k-1) \leq n-2 \implies s_k \leq \dfrac{n-2}{k-1} \implies s_k \leq \left\lfloor\dfrac{n-2}{k-1} \right\rfloor$$ This bound is tight, i.e., you can construct a tree with exactly $\left\lfloor\dfrac{n-2}{k-1} \right\rfloor$ vertices of degree $k$.
Hence, in the case of $k=3$, we have $s_3 \leq \left\lfloor\dfrac{n}2 \right\rfloor - 1$.
I hope I understood your question correctly.