In any tree, what is the maximum distance between a vertex of high degree and a vertex of low degree? In any undirected tree $T$, what is the maximum distance from any vertex $v$ with $\text{deg}(v) \geq 3$ to the closest (in a shortest path sense) vertex $y$ with $\text{deg}(y) \leq 2$? That is, $y$ can be leaf.
It seems to me that this distance can be at most $\dfrac{\text{diam}(T)}{2}$, and furthermore that the maximum distance will be attained from a graph center. Is this true? There's probably simple argument for it somewhere.
 A: In answer to the first version of the question (between any $v$ and any $y$):
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y is at the left, v is at the vee
The answer to the second version of the question is yes.  Suppose that $v$ were more than $\frac{D}{2}$ away from all leaves, where $D$ denotes the graph diameter.  Pick two of the branches leading from $v$.  Each of them contains at least one leaf.  The shortest path between these leaves passes through $v$, and is of distance more than $D$, being the sum of two distances each more than $\frac{D}{2}$.  This is a contradiction.
The bound given is tight, by considering the Y graph on four vertices.
A: In your second question (shortest path to vertex of degree $\le 2$), the bound $\operatorname{diam}(G) / 2$ holds, simply by noticing that the ends of the longest path in a tree are leaves, and the "worst" that a graph can do is have the vertex of degree $3$ or higher right in the center.  But in fact, this holds for the shortest path from any vertex of to vertex of degree $1$.
Can we do better?  No, in fact.  Take a look at a tournament bracket, for example (just delete the leaves on one side to create a unique center).  Viewed as a tree, all the vertices are of degree $3$ except for the leaves, which are all at least distance $d/2$ from the center of this modified graph.
