Let $A \in M_{m\times n}(\mathbb{Z})$, then $A$ is totally unimodular iff for any $B$, square submatrix of $A$, then $|\det(B)| \leq 1$. If $A$ is itself a square matrix, then we need also $|\det(A)|=1$.
My question will be the "combinatorially" following: "How many totally unimodular matrices are there?"
It's well known the upper bound of $3^{nm}$ of all $(0,1,-1)$-entries matrices and the lower bound of $2\max({ n \choose 2}^m,{m \choose 2}^n)$ of consecutive plus/minus-ones property in each row/column. But I don't have any significantly better approach.