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Let $G=(V,E)$ be an undirected tree.

Let $d$ be the maximum degree of a vertex in the tree.

Prove that $G$ has at least $d$ leaves.

I need some help proving this, should I consider a run of BFS on that vertex?

Is that the correct direction?

Thanks!

RobPratt
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CalculusLover
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1 Answers1

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When you cut the tree at the given vertex of maximal degree $d$, you get $d$ separate trees. Show that each such tree has at least two leaves. One of them can possibly be the vertex at which you cut the original tree, but even if you take that into account, you still get $d$ leaves. Show that these leaves are also leaves in the uncut graph.

I leave the details for you to figure out.

GSofer
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  • That sounds good, thanks! A quick question - by saying "cut the tree at the given vertex" - do you include this vertex? If yes then I get $d+1$ trees, and if no I get $d$ trees. Thanks! :) – CalculusLover Feb 20 '22 at 16:42
  • @CalculusLover You "create" $d$ copies of the given vertex, so that it is included in each individual subtree. For instance, if you do this procedure with the main vertex of a star graph, you get $d$ intervals. – GSofer Feb 20 '22 at 20:20