TITLE: Graphs in which every spanning tree is an independency tree
QUESTION [12 upvotes]: It follows from this question 
and the corresponding answers, that the complete graphs and the cycles are precisely the graphs 
$G$ having the property that, for every spanning tree $T$ of $G$, the set of leaves of $T$ is a 
clique in $G$. 
Motivated by this fact, I am looking for a characterization of all (connected) graphs $G$ 
having the property that, for every spanning tree $T$ of $G$, the set of leaves of $T$ is an 
independent set in $G$. 
(Such spanning trees are called independency trees 
in the literature.)

REPLY [3 votes]: Here's something more about these graphs.

Theorem: Let $G$ be connected graph with $\geq 3$ vertices. Consider the next properties:
1) Every edge in $G$ is incident to a cut vertex.
2) $G$ has all spanning trees independency.
3) The set of all cut vertices in $G$ is dominating.
Then 1)$\Rightarrow$2)$\Rightarrow$3).

Proof 
1)$\Rightarrow$2): Let $e=uv\in E(G)$ and $u$ is a cut vertex in $G$. Assume that $v$ isn't a leaf. Then $G-u$ is disconnected and $d_{G-u}(v)\geq 1$. Therefore $G-\{u,v\}$ is also disconnected. Thus $\{u,v\}$ is a separator in $G$.
2)$\Rightarrow$3): Assume that for some non-cut vertex $u\in V(G)$ the set $N(u)$ consists of non-cut vertices. Then for all $v\in N(u)$ the set $\{u,v\}$ is a separator in $G$. It means that $v$ is a cut vertex in $G-u$.
If there exists a component $A$ of $G-\{u,v\}$ such that $V(A)\cap N(u)=\emptyset$, then $A$ is also a component of $G-v$. It means that $v$ is a cut vertex in $G$ which is a contradiction. Thus every component of $G-\{u,v\}$ contains the vertices from $N(u)$. 
Therefore, each vertex $v\in N(u)$ separates the set $N(u)-v$ in $G-u$. But this contradicts the connectedness of $G-u$. $\boxtimes$
Two counterexamples to the reverse statements of the Theorem.
2)$\nRightarrow$1): 

3)$\nRightarrow$2): $K_{1}+(P_{2}\cup K_{1})$.