Let $T$ and $T'$ be 2 distinct spanning trees of a connected graph of order $n$. Show that there exists a sequence $T=T_0,T_1,\dots,T_k=T'$ of spanning trees of $G$ such that $T_i$ and $T_{i+1}$ have $n-2$ edges in common for each $i$ such that $0 \leq i\leq k-1$
I know that $T$ and $T'$ are 2 distinct spanning trees of a connected graph of order $n$, so both of them has order $n$ and size $n-1$. I also know that in order for $T$ and $T'$ to be distinct, there must be at least one edge $e \in E(T)$ but $e \not \in E(T')$ and one edge $e' \in E(T')$ but $e' \not \in E(T)$. Since both $T$ and $T'$ have $n-1$ edges, $T$ and $T'$ must have at most $n-2$ edges in common,.
However, I'm kinda confused about the sequence part.
The result essentially says that you can change $T$ into $T'$ by changing one edge at a time in such a way that you'll always have a tree.
If I were to do this with words I would get something like this:
POT COT CON CAN TAN TAP TOP
(Obviously there are much quicker ways of getting from POT to TOP.)
My first thought was to take away an edge of $T$ and add an edge of $T',$ but it turned out to be easier to do it the other way round:
Starting with $T$, add an edge of $T'$ that isn't already in $T$ (we can do this because the trees are distinct). This creates a cycle. $T'$ is acyclic, so at least one edge of the cycle isn't in $T'.$ Pick any such edge and remove it from the graph. This gives us a new spanning tree. Then you just repeat this process until you get to $T'.$