## 3 \(\gamma\)-Trees of Trees
For any component, \(T^{\prime}\) of \(T-N[x]\), \(D\cap V(T^{\prime})\) is the unique \(\gamma\)-set of \(T^{\prime}\). Therefore \(D\) is the unique \(\gamma\)-set of \(T\) containing \(x\). Theorem 4.1: _If \(G\) is a \(\gamma\)-graph of a tree \(T\) and \(L\) is a leaf of \(G\), then \(G-L\) is a \(\gamma\)-graph of some tree \(T^{\prime}\)._  
Proof: By Lemma 4, exactly one vertex \(x\) in \(L\) has fewer than two \(L\)-external private neighbours. Moreover, either \(x\) has exactly one \(L\)-external private neighbour or \(x\) is a leaf in \(T\) and an \(L\)-self private neighbour. By Lemma 4.1, \(L\) is the unique \(\gamma\)-set of \(T\) containing \(x\). If \(x\) is a leaf of \(T\) and is an \(L\)-self private neighbour, then \(T^{\prime}\) is formed by adding a leaf to the stem of \(x\). Suppose on the other hand, \(x\) has exactly one external private neighbour \(y\). If \(x\) is a self private neighbour, then \(T^{\prime}\) is formed by rooting \(T\) at \(y\), deleting the descendants of \(x\), and adding a leaf to \(y\). Otherwise, \(T^{\prime}\) is formed by rooting \(T\) at \(y\), and deleting \(x\) and its descendants. Corollary 2: _If \(G\) is a tree and a \(\gamma\)-graph of a tree \(T\), then every subtree of \(G\) is a \(\gamma\)-graph of some tree._  
Corollary 3: _If \(G\) is a tree and a \(\gamma\)-graph of a tree \(T\), then \(\mathbf{H}\) is not a subtree of \(G\)._  
Proof: The result immediately follows from Lemma 4.1 and Corollary 2. From the previous corollary, we know that if a tree is a \(\gamma\)-graph of a tree, then the vertices of degree at least three are not adjacent. We now wish to show all trees without adjacent vertices of degree at least three are the \(\gamma\)-graph of a tree. The proof is constructive. We establish first some building blocks. Theorem 4.2: _Let \(Y_{n}\) be the graph obtained by taking \(n\) copies of \(K_{1,3}\) and joining a leaf of each copy to a common vertex (see Figure 2 for an example). Then \(Y_{n}(\gamma)\cong K_{1,n}\), and in every \(\gamma\)-set which is a leaf of \(K_{1,n}\), every vertex has an external private neighbour._  
Proof: Since every stem of \(Y_{n}\) is adjacent to two leaves, every \(\gamma\)-set of \(Y_{n}\) must contain every stem. Only the common centre vertex is left to be dominated, so we can form a \(\gamma\)-set by adding this vertex or any of its neighbours. Further, the only edges are between the \(\gamma\)-set containing the common centre vertex and each of the other \(\gamma\)-sets. Finally, in each of these other \(\gamma\)-sets, every stem has at least two external private neighbours, its leaves, and the additional vertex has the common centre vertex as its private neighbour. The result follows. We now provide a tool for combining these building blocks. Figure 2: The graph \(Y_{3}\). **Lemma 9**: _Suppose for \(i=1,2\), \(G_{i}\) is a \(\gamma\)-graph of tree \(T_{i}\), \(X_{i}\) is a leaf of \(G_{i}\) and every vertex in the dominating set \(X_{i}\) has at least one \(X_{i}\)-external private neighbour. By Lemma 4, in each \(X_{i}\) there is exactly one vertex \(v_{i}\in V(T_{i})\) with fewer than two \(X_{i}\)-external private neighbours. Then \(G\), the graph formed by identifying \(X_{1}\) and \(X_{2}\) is the \(\gamma\)-graph of a tree \(T:\cong T_{1}\oplus_{2_{2}}^{v_{1}}T_{2}\), the tree formed by linking \(T_{1}\) and \(T_{2}\) by creating a new vertex \(v\) adjacent to \(v_{1}\) and \(v_{2}\)._  
_Furthermore let \(Y\in V(G_{1})\) be a dominating set of \(T_{1}\) and \(Y^{\prime}\in V(G)\) be the corresponding dominating set of \(T\). If every vertex of \(Y\) has a \(Y\)-external private neighbour, then every vertex in \(Y^{\prime}\) has a \(Y^{\prime}\)-external private neighbour._  
By Lemma 4, exactly one vertex \(v_{i}\) in \(X_{i}\) has fewer than two external private neighbours. It follows that \(v_{i}\) has exactly one external private neighbour. Form \(T\) from \(T_{1}\) and \(T_{2}\) by adding a new vertex \(v\) with \(N(v)=\{v_{1},v_{2}\}\). We now show that \(T(\gamma)\cong G\).