Let $G$ be a connected simple graph. For any spanning tree $T$ of $G$, let $l(T)$ be the number of leaves of the graph $T$. Consider $\ell=\min\_Tl(T)$, can I find a spanning tree $T$ with $l(T)=\ell$, such that the set of leaves $A$ of $T$ is very close to an independent set.


For example, I guess that there exists a vertex $a\in A$ such that $A\setminus\{a\}$ is an independent set in the graph $G$.


My idea is that maybe we can justifying the tree $T$ step by step to get a extreme tree with some properties.


Is there any results or references for this question?



Property of the spanning tree with minimal leaves
If I am not mistaken, and if I understand you correctly, it seems to me that you are right.


The following statement is true.



> 
> Let $G$ be a connected graph and $T$ be the spanning tree with the
> smallest number of leaves and $\ell=l(T)$. Let $A$ be the set of all
> leaves of tree $T$. If $|A|=\ell>2$, then $A$ is an independent set of
> graph $G$.
> 
> 
> 


Here is a brief proof. Let $x$ and $y$ be two leaves and $e=xy$ be an edge of graph $G$.
Then the graph $H=T+e$ has a cycle. Denote this cycle by $C$. If all vertices of the cycle $C$ have degree $2$ in $H$, then $C=H$ and our graph $G$ is Hamiltonian, and this contradicts the condition $\ell>2$.


Hence there exists a vertex $a$ of cycle $C$ of degree $3$ or more in $H$.
Let $e'=ab$ be an edge of $C$. The graph $T'=H-e'$ is a spanning tree of graph $G$ and $l(T')<\ell$. Contradiction.


