If a tree on n vertices (n>1) has 2 pendant vertices, does it have Euler path or Hamilton path?
2026-03-26 01:26:38.1774488398
Tree & Euler, Hamilton paths
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If a tree has 2 pendant vertices, then it must be a path (and therefore has an Eulerian path and a Hamiltonian path).
To see this, let $T$ be a tree which is not a path. Choose a maximal path, $P$ that is a subgraph of $T$. Its endpoints are each pendant vertices, so there are at least 2 pendant vertices. Since $T$ is not a path, there exists at least one vertex which is not in $P$, but is adjacent to a vertex in $P$. If it is of degree one, then $T$ has more than 2 pendant vertices. If it is not of degree one, then it must be adjacent to a vertex which is not in $P$ (since trees are acyclic). Following this process with the newfound vertex, it must either be pendant or adjacent to another vertex which is not in $P$. After a finite number of steps continuing this, we must eventually reach a pendant vertex, so $T$ has more than 2 pendant vertices.