Finite Tree has Leaf Nodes

Theorem
Every non-null finite tree has at least two nodes of degree $$1$$.

Proof
We use the Method of Infinite Descent.


 * Suppose $$T$$ is a tree which has no nodes of degree $$1$$.

First note that no tree has all nodes of even degree.

That is because it would then be an Eulerian Graph, and by definition, trees do not have circuits.

From the Handshake Lemma, we know that $$T$$ must therefore have at least two nodes whose degree is odd and at least $$3$$.

As $$T$$ is finite, this number must itself be finite.

Of those nodes, there must be two (call them $$u$$ and $$v$$) which can be joined by a path $$P$$ containing no odd nodes apart from $$u$$ and $$v$$.

(Otherwise you can pick as one of the two nodes one of those in the interior of $$P$$.)

Consider that path $$P$$ from $$u$$ to $$v$$.

As a tree has no circuits, all nodes of $$P$$ are distinct, or at least part of $$P$$ will describe a cycle.

Now consider the subgraph $$S$$ formed by removing the edges comprising $$P$$ from $$T$$, but leaving the nodes where they are.

The nodes $$u$$ and $$v$$ at either end of $$P$$ will no longer be odd, as they have both had one edge removed from them.

All the nodes on $$P$$ other than $$u$$ and $$v$$ will stay even.

The graph $$S$$ may become disconnected, and may even contain isolated nodes.

However, except for these isolated nodes (which became that way because of being nodes of degree $$2$$ on $$P$$), and however many components $$S$$ is now in, all the nodes of $$S$$ are still either even or odd with degree of $$3$$ or higher.

That is because by removing $$P$$, the only odd nodes we have affected are $$u$$ and $$v$$, which are now even.

Now, if the nodes in any component of $$S$$ are all even, that component is Eulerian.

Hence $$S$$ contains a circuit, and is therefore not a tree.

From Subgraph of Tree, it follows that $$T$$ can not be a tree after all.

However, if the nodes in any component $$T'$$ of $$S$$ are not all even, then there can't be as many odd nodes in it as there are in $$T$$ (because we have reduced the number by $$2$$).

Also, because of the method of construction of $$T'$$, all of its odd nodes are of degree of at least $$3$$.

So we repeat the above construction on $$T'$$, and obtain:
 * either a subgraph with a component whose nodes are all even (in which case it is Eulerian;
 * another subgraph $$T''$$ all of whose odd nodes are of degree of at least $$3$$.

Each time we perform the construction, we reduce the number of such odd nodes by $$2$$, and never do we create a node with degree $$1$$.

Sooner or later we will reach a subgraph of $$T$$ all of whose nodes are even.

Thus we see that we can not avoid the fact that $$T$$ contains an Eulerian subgraph.

So $$T$$ contains a circuit, and is therefore not a tree.

So $$T$$ must have at least one node of degree $$1$$.


 * Now, suppose that $$T$$ is a tree with exactly $$1$$ node of degree $$1$$. Call this node $$u$$.

From the Handshake Lemma, we know that $$T$$ must therefore have at least one node whose degree is odd and at least $$3$$.

Let $$P$$ be the path from $$u$$ to any such odd node that passes only through even nodes, as we did above.

Again, let us remove all the edges of $$P$$.

By the argument above, we will once again have reduced the tree to one in which any remaining odd nodes all have degree of at least $$3$$.

Then we are in a position to apply the argument above.

Hence $$T$$ must have at least two node of degree $$1$$.

Note
It is instructive to see what happens to the above argument for a tree with exactly two nodes of degree $$1$$.

There may be no other nodes of odd degree but these two (call them $$u$$ and $$v$$).

Such a graph is a path graph, and removing all the edges from the path from $$u$$ to $$v$$ leaves a null graph, which of course has no circuits.