# Show that any tree $T$ has at least $\Delta(T)$ leaves. Is my proof correct?

I'm attempting exercise 15 from the end of the first section in Diestel's Graph Theory. The exercise asks:

Show that any tree $T$ has at least $\Delta(T)$ leaves. $\Delta(T)$ denotes the maximum degree of any vertex in T. $d(v)$ denotes the degree of vertex $v$

I came up with the following:

$Proof.$ Let $T = (V, E)$ be a tree with an arbitrarily selected root, $r$. Let $L \subseteq V$ denote the set of vertices of degree 1 in T, i.e. the leaf vertices of $T$.

Now let $v \in V - L$ be an inner-vertex of $T$ (i.e. not a leaf), and let $d(v) = \Delta(T)$. As an inner-vertex of the tree, clearly $v$ must lie on some path $rTu$ for some $u \in L$, and as a result, $1$ of the $\Delta(T)$ incident edges of $v$ must lead to $u$. By the definition of a tree, each path from $r$ to a leaf $w \in L$ is uniquely defined. It follows that all $\Delta(T) - 1$ other edges incident to $v$ must belong to unique paths leading to other leaf nodes $w \neq u$, giving us at least $\Delta(T)$ leaves. $\Box$

I'm not convinced. Any suggestions?

I would structure the proof as follows. Consider an internal vertex $v$ with degree $\Delta(T)$; we would like to show that there are at least $\Delta(T)$ leaves that are descendants of $v$. Each child of $v$ is either a leaf or an internal vertex. Suppose there are $n_1$ children that are leaves and $n_2$ that are internal vertices. We have $n_1 + n_2 = \Delta(T)$.

Each child $c$ of $v$ that is an internal vertex will only have leaf descendants that are not descendants of any other child $c'$ of $v$. For if they did indeed have a common descendant $d$, there would be more than one path from $d$ to $v$. Thus there are at least $n_2$ such leaves, and consequently there are at least $n_1 + n_2 = \Delta(T)$ leaves.

• Thanks for your response. I agree, your proof is far more explicit, and reads better. Just for clarification, was the general gist of my attempt correct? – swingballchamp42 Oct 3 '17 at 17:20
• The gist of your attempt is correct (you are making use of the same uniqueness property that I mention). – Hans Hüttel Oct 3 '17 at 17:33

Now let $v\in V−L$ be an inner-vertex of $T$ (i.e. not a leaf), and let $d(v)=\Delta(T)$

When you say "let $v\in V-L$", it sounds like you are talking about an arbitrary vertex, and when you say "let $d(x)=\Delta(T)$", it sounds like you want $v$ to be a particular vertex, namely the one of largest degree. So you should replace the first quotation with "let $v$ be the vertex in $T$ with the largest degree".

It follows that all $Δ(T)−1$ other edges incident to vv must belong to unique paths leading to other leaf nodes $w≠u$.

I think this conclusion is where you are sweeping details under the rug...

Here's how I would structure the proof. You do not need to talk about the root of the tree at all.

Consider a vertex $v$ which achieves the maximum degree $\Delta(T$), and consider what happens when you remove $v$. Show that

• the resulting graph has several connected components, each of which is a tree
• the neighbors of $v$ are all in different components,
• each component has two leaves (this is a standard theorem, every tree has at least two leaves)
• at least one of the two leaves in each component corresponds to a leaf in the original graph $T$.