## 3 \(\gamma\)-Trees of Trees
It is clear that \(\gamma(T)\leq\gamma(T_{1})+\gamma(T_{2})\). Suppose some \(\gamma\)-set \(D\) of \(T\) contains \(v\). Then without loss of generality, \(\gamma(T_{1}-\{v_{1}\})<\gamma(T_{1})\). Let \(D^{\prime}\) be a \(\gamma\)-set of \(T-\{v_{1}\}\). Then no neighbour of \(v_{1}\) is in \(D^{\prime}\). Hence \(D^{\prime}\cup\{v_{1}\}\) is a \(\gamma\)-set of \(T_{1}\) and \(v_{1}\) has no external private neighbours. Then, by Lemma 8, \(D^{\prime}\cup\{v_{1}\}\) is the unique \(\gamma\)-set of \(T_{1}\) containing \(v_{1}\), so \(D^{\prime}\cup\{v_{1}\}=X_{1}\). But this contradicts the presupposition that every vertex in \(X_{1}\) has an \(X_{1}\)-external private neighbour. Hence no \(\gamma\)-set of \(T\) contains \(v\), \(\gamma(T)=\gamma(T_{1})+\gamma(T_{2})\) and every \(\gamma\)-set of \(T\) is the union of a \(\gamma\)-set of \(T_{1}\) and a \(\gamma\)-set of \(T_{2}\). Every \(\gamma\)-set of \(T\) must contain \(v_{1}\) or \(v_{2}\) for \(v\) to be dominated. Let \(X=X_{1}\cup X_{2}\). It is clear that \(X\) is the unique \(\gamma\)-set containing both \(v_{1}\) and \(v_{2}\). Since every other vertex in \(X\) has least two external private neighbours, and \(v_{1}\) and \(v_{2}\) have exactly one external private neighbour, \(X\) is adjacent to two \(\gamma\)-sets. Let \(D\) be a dominating set of \(T\) other than \(X\). If \(v_{2}\in D\) and by Lemma 8, \(D\cap T_{2}\cong X_{2}\) and every vertex in \(D\cap T_{2}\) has at least two \(D\)-external private neighbours. Therefore the subgraph of the \(\gamma\)-sets of \(T\) containing \(v_{2}\) in \(T(\gamma)\) is \(G_{1}\) and similarly the subgraph of the \(\gamma\)-sets of \(T\) containing \(v_{1}\) in \(T(\gamma)\) is \(G_{2}\). Hence, the \(\gamma\)-graph of \(T\) is \(G\) as required. The final statement follows directly from the above construction. We can now prove the main result. **Theorem 10**: _If \(G\) is a tree, then \(G\) is a \(\gamma\)-graph of some tree if and only if \(H\) is not a subtree of \(G\)._  
Necessity follows by Corollary 3; it remains to show sufficiency. The result trivially holds for \(K_{1}\) as \(K_{1}(\gamma)\cong K_{1}\); so suppose \(G\) has at least two vertices. We proceed by induction on \(k\), the number of degree two vertices of \(G\). If \(k=0\), then \(G\cong K_{1,n-1}\), and by Theorem 9, \(K_{1,n-1}\) is the \(\gamma\)-graph of \(Y_{n-1}\), and in every \(\gamma\)-set which is a leaf of \(K_{1,n-1}\), every vertex has an external private neighbour. Assume \(k>0\) and that every tree \(F\) not containing \(H\) as a subtree with fewer than \(k\) degree two vertices is the \(\gamma\)-graph of some tree and in every \(\gamma\)-set which is a leaf of \(F\), every vertex has an \(F\)-external private neighbour. Let \(x\) be a degree two (cut-)vertex in \(G\), and let \(G_{1}\) and \(G_{2}\) be the two components of \(G-\{x\}\). As \(G_{1}\cup\{x\}\) and \(G_{2}\cup\{x\}\) each have fewer than \(k\) degree two vertices, then by the induction hypothesis, each is the \(\gamma\)-graph of a tree (say \(T_{1}\) and \(T_{2}\) respectively) and in every \(\gamma\)-set which is a leaf of \(T_{1}\) and \(T_{2}\) repsectively, every vertex has an external private neighbour. By Lemma 9, there are vertices \(v_{1}\in V(T_{1})\) and \(v_{2}\in V(T_{2})\) so that the \(\gamma\)-graph of \(T_{1}\oplus_{p_{2}}^{v_{1}}T_{2}\) is \(G\) and by construction, in every \(\gamma\)-set which is a leaf of \(G\), every vertex has an external private neighbour. The result follows.