Tree has Center or Bicenter

Theorem
Every tree has either:


 * Exactly one center, or:


 * Exactly one bicenter,

but never both.

That is, every tree is either central or bicentral.

Proof
A tree whose order is $$1$$ or $$2$$ is already trivially central or bicentral.

Let $$T$$ be a tree of order at least $$3$$.

First we establish that the construction of a center or bicenter actually works.

From Tree has Degree One Nodes, there are always at least two nodes of degree $1$ to be removed.

By the Handshake Lemma, it is clear that $$T$$ must also have at least one node whose degree is greater than $$1$$:


 * $$\sum_{i=1}^n \deg_G \left({v_i}\right) = 2q$$

where $$q$$ is the number of edges in $$T$$.

But $$q = n-1$$ from Number of Edges in Tree.

So if each node has degree $1$, then $$n = 2 \left({n-1}\right)$$ and so $$n = 2$$.

Therefore if $$n > 2$$ there must be at least one node in $$T$$ of degree is greater than $$1$$.

Next, from Subgraph of Tree, after having removed those nodes, what is left is still a tree.

Therefore the construction is valid.

We need to show the following:


 * 1) $$T$$ has only one center or bicenter;
 * 2) $$T$$ has either a center or a bicenter.


 * Suppose $$T$$ has more than one center or bicenter.

It would follow that at least one of the iterations constructing the center or bicenter disconnects $$T$$ into more than one component.

That could only happen if we were to remove an edge between two nodes of degree greater than $$1$$.

Hence $$T$$ has at most one center or bicenter.


 * Now to show that $$T$$ has at least one center or bicenter.

The proof works by the Principle of Strong Induction.

We know that a tree whose order is $$1$$ or $$2$$ is already trivially central or bicentral. This is our base case.

Suppose that all tree whose order is $$n$$ have at most one center or bicenter. This is our induction hypothesis.

Take a tree $$T$$ whose order is $$n+1$$ where $$n > 2$$.

Let $$T$$ have $$k$$ nodes of degree $1$.

We remove all these $$k$$ nodes.

This leaves us with a tree with $$n+1-k$$ nodes.

As we have seen that $$T$$ has at least one node whose degree is greater than $$1$$, $$n+1-k \ge 1$$.

As there are always at least two nodes of degree $1$, $$n+1-k \le n-1$$.

So after the first iteration, we are left with a tree whose order is between $$1$$ and $$n-1$$ inclusive.

By the induction hypothesis, this tree has either a center or bicenter.

The result follows by the Principle of Strong Induction.