## 3 \(\gamma\)-Trees of Trees  
In this section, we characterize the trees which are \(\gamma\)-graphs of trees. We first consider properties of the \(\gamma\)-sets corresponding to the leaves of a \(\gamma\)-graph. (Recall we equate each node in \(G(\gamma)\) with its corresponding \(\gamma\)-set of \(G\).) **Lemma 4**. : _Let \(T\) be a tree with at least three vertices and let \(T(\gamma)\) be the \(\gamma\)-graph of \(T\). If \(D\) is a leaf in \(T(\gamma)\), then exactly one vertex \(v\in D\) has fewer than two \(D\)-external private neighbours. Furthermore either \(v\) has exactly one \(D\)-external private neighbours or \(v\) is both a leaf in \(T\) and a \(D\)-self private neighbour._  
Proof. : Note that if every vertex of dominating set \(D\) in a tree has at least two \(D\)-external private neighbours, \(D\) has no neighbours in \(T(\gamma)\). If a vertex \(x\) in \(T\) has exactly one external private neighbour \(y\) in \(D\), then \((D-\{x\})\cup\{y\}\) is a \(\gamma\)-set of \(T\) adjacent to \(D\). If a vertex \(z\) in \(T\) has no external private neighbours in \(D\), then \(z\) must be a self private neighbour in \(D\) and for every \(w\in N(z)\), \((D-\{z\})\cup\{w\}\) is a \(\gamma\)-set of \(T\) adjacent to \(D\). As \(D\) is adjacent to exactly one \(\gamma\)-set, the result follows. Next, we prove a fundamental result on the number of private neighbours of a vertex in adjacent \(\gamma\)-sets of a tree. **Lemma 5**. : _Let \(D\) and \(F\) be \(\gamma\)-sets of a tree \(T\) which are adjacent in \(T(\gamma)\) and let \(x\in D\cap F\). Then \(|pn(x,D)\setminus pn(x,F)|\leq 1\) and \(|pn(x,F)\setminus pn(x,D)|\leq 1\)._  
Proof. : Suppose \(F=(D-\{y\})\cup\{z\}\) for adjacent vertices \(y\) and \(z\) in \(T\). Let \(x\in D\cap F\). Suppose the result is false and assume without loss of generality that \(|pn(x,D)\setminus pn(x,F)|\geq 2\). If \(x\) is a \(D\)-self private neighbour, but not an \(F\)-self private neighbour, then \(z\sim x\) and \(z\) is adjacent to a \(D\)-external private neighbour of \(x\), say \(w\). Then of \(xwz\) is a \(3\)-cycle in \(T\), which contradicts that \(T\) is a tree. Otherwise, \(x\) has at least two \(D\)-external private neighbours \(v,w\) that are adjacent to \(z\), so \(xvzw\) is a \(4\)-cycle in \(T\), which contradicts that \(T\) is a tree. The result follows. We now consider the \(\gamma\)-sets of leaves adjacent to the same stem. **Lemma 6**. : _Let \(T\) be a tree and let \(T(\gamma)\) be the \(\gamma\)-graph of \(T\). If \(S\) is a stem in \(T(\gamma)\) with degree at least three and \(L_{1},L_{2},\ldots,L_{k}\) are the leaves of \(S\), then there exists vertices \(x\in S\) and \(y_{1},y_{2},\ldots,y_{k}\in V(T)-S\) so that \(x\sim y_{i}\) and \(L_{i}=(S-\{x\})\cup\{y_{i}\}\) for \(i=1,2,\ldots k\)._  
Proof. : If \(k=1\), the statement is trivial, so assume \(k\geq 2\). As \(S\) is adjacent to \(L_{i}\) in \(T(\gamma)\), for each \(i\) there exists vertices \(x_{i}\) and \(y_{i}\) so that \(L_{i}=(S-\{x_{i}\})\cup\{y_{i}\}\) and \(x_{i}\sim y_{i}\). Let \(X=\{x_{i}:i=1,2,\ldots,k\}\subseteq S\)Suppose \(|X|\geq 2\). By Lemma 4, \(y_{i}\) is the only vertex in \(L_{i}\) with fewer than two \(L_{i}\)-external private neighbours. In particular, if \(x_{i}\
eq x_{j}\), then \(x_{i}\) has at least two \(L_{j}\)-external private neighbours. It now follows from Lemma 5, that each \(x_{i}\) has an \(S\)-external private neighbour. If \(x_{i}\
eq x_{j}\), then \(S-\{x_{i},x_{j}\}\cup\{y_{i},y_{j}\}\) is not a \(\gamma\)-set of \(T\) as \(L_{i}\) and \(L_{j}\) are leaves of \(T(\gamma)\). Hence there exists a vertex \(z_{ij}\) which is not dominated by \(S-\{x_{i},x_{j}\}\cup\{y_{i},y_{j}\}\). It must be the case that both \(x_{i}\) and \(x_{j}\) are neighbours of \(z_{ij}\). If \(|X|\geq 3\), then each such \(z_{ij}\) must be identical or \(T\) is not a tree, but then \(S-\{x_{i},x_{j}\}\cup\{y_{i},y_{j}\}\) is dominating, a contradiction. Hence we assume \(|X|=2\), say \(X=\{x_{1},x_{2}\}\).