Any tree with $|V|\geq 2$ has at least two pendant (degree 1) node. **Proof**: If $|V|=2$ then $G$ must be isomorphic to $(\{ u,v \}, \{ \{ u,v \} \})$ which has certainly has two pendant nodes. Otherwise, suppose $|V|\geq 3.$ Since a tree has $n-1$ edges, by handshaking lemma, the degree sum is $2(n-1).$ However, if no node has degree greater than or equal to $2$ then the degree sum is strictly less than $n$ which gives a contradiction. Thus, there exists a node with degree greater than two. Therefore by [[Tree with degree k node has at least k Pendant nodes]], the tree has at least 2 pendant nodes.