For the sake of completeness here's the proof of McKay's characterization. By $L(G)$ I denote the set of all leaves in $G$.

**Theorem:** Every spanning tree in $G$ is independency tree if and only if every edge $e=uv\in L(G)$ is incident to a leaf or the set $\{u,v\}$ is a separator in $G$.

*Proof:* $\Rightarrow$ Assume that for some edge $e=uv\in E(G)$ the set $\{u,v\}$ isn't a separator and $u,v\notin L(G)$. Then the graph $H=G-\{u,v\}$ is connected.

Now fix some spanning tree $T'\subset H$.

Since $u,v\notin L(G)$, it holds that $A=N_{G}(u)\cap V(T')\neq\emptyset$ and $B=N_{G}(v)\cap V(T')\neq\emptyset$. Choose $w_{1}\in A$, $w_{2}\in B$ and put $$V(T)=V(G),\ E(T)=E(T')\cup\{uw_{1},vw_{2}\}.$$
Clearly, $T$ is a spanning tree in $G$ and $u,v\in L(T)$ with $uv\in E(G)$ which is a contradiction.

$\Leftarrow$ Let $T\subset G$ be some spanning tree and $u,v\in L(T)$. Clearly, $G-\{u,v\}$ is connected. Therefore the set $\{u,v\}$ never a separator.

Again, assume that $e=uv\in E(G)$. Then $d_{G}(u)\geq 2$ and $d_{G}(v)\geq 2$. It means that $u,v\notin L(G)$. Again, a contradiction.

properedge from the cycle passing through $uv$ and remove $uv.$ One then obtains a spanning tree in which $u$ and $v$ are leaves. There is some technicality in assuring the chosen edge is not incident with $u$ and $v$ though. $\endgroup$