Prove that if a graph contains a leaf vertex then it is not Hamiltonian? I need to prove that if a graph contains a leaf vertex then it is not Hamiltonian.
I get that it wouldn't be Hamiltonian because to be Hamiltonian you have to have a Hamiltonian cycle and if you have a leaf vertex there's no possible way to have a cycle including that vertex. I'm just not sure that this alone is a valid argument. 
 A: First the definition of Hamiltonian cycle is that it visits each node of the graph exactly once.
Now, the proof can be done by contradiction:
Suppose there is a Hamiltonian graph and our leaf node is X. So it is connected to only one other node, say Y.
Let's start traversing the graph along the existing Hamiltonian cycle. We can start from any node so assume that it wasn't X. At the point when we visit X we must have come from Y but X is not closing the cycle so we need to continue our traversal. But the only possibility is to go back to Y since X is a leaf node, which gives us the contradiction.   
A: You seem to understand the idea, although "...if you have a leaf vertex there's no possible way to have a cycle including that vertex" is imprecise.  This can be made more precise by describing the vertex degrees.
If I were asked to prove this, I'd instead prove the stronger result:

Any graph $G$ with a $2$-regular spanning subgraph $H$ has no leaf vertices.

By definition, every vertex in $H$ has degree $2$, and $H$ is a subgraph of $G$.  This gives a lower bound on the minimum vertex degree in $G$.  If $G$ had a leaf  vertex, it would violate this lower bound.
The desired result follows since a Hamiltonian cycle is an example of a $2$-regular spanning subgraph.
