Does an algorithm exist that transforms any connected graph, cyclic or not, into tree form? I developed an algorithm that transforms any simple connected graph, cyclic or not, into a tree.
The resulting tree is syntax-preserving, in a sense that it allows to reconstruct the original input graph and only the original input graph. In other words, the constructed tree preserves adjacency information, while resolving cycles.
Moreover, I assume that with the constructed tree allowing to reconstruct only the original input graph, no other graph that is structurally different can exist that would result in an identical tree form. 

I did some research on algorithms that transform graphs into tree form, but I was unable to find another algorithm that would work for any simple connected graph like described above.
However, I am pretty such there must exist something. 

So maybe one more experienced in graph theory can help me out.
It would be highly appreciated. 

Alex
In a general graph, the number of edges can go up to $\mathcal{O}(n^2)$ (where n is the number of verticies). But in a tree, that number is always $n-1$.

Then, we understand that its impossible to convert a graph to a tree preserving syntax without increasing the memory for each node by a linear factor, or increasing the number of nodes significantly.

Let us define a unique number for every node (in any way you like).

The algorithm now will be:


Let $\mathcal{T}$ be the tree from the BFS algorithm on the graph (starting from node 1, for uniqueness)
For every edge $e=(i,j)$ in the graph, that is not in $\mathcal{T}$ already, w.l.o.g assume $i\le j$, and add a new node with value $j$ to $\mathcal{T}$, and a connection between it and the (original) node $i$ in the tree.


Note that in the new tree, there will be a lot of nodes that have the same number.

To reconstruct the original graph, we will do the following:


Create a node for every unique number in $\mathcal{T}$.
Add an edge $(i,j)$ if there is a node with number $i$ that has an edge to some node with value $j$