Technically, they are. There was a post on HN recently about how every model is equivalent to a decision tree.

I mean, if it can be specified by a computational process, that’s trivially true, right?

I'm not so sure about that.

Decision trees are finite set of if-else branches, and work on finite input - much like models. I assume many of the classic arbitrary-size-input algorithms cannot be represented by a decision tree.

Adding two arbitrary-sized numbers comes to mind - how would one model it as a decision tree?

    if (a[0] + b[0] == 0) {
        c[0] = 0
    } else if (a[0] + b[0] == 1) {
        c[0] = 1
    } else if ...
    ...
    } else if (a[0] + b[0] == 10) {
        c[0] = 0
        if (a[1] + b[1] == 0) {
            c[1] = 1
        } else ...
        ...
    }
    ...

You can see that for arbitrary-sized input, this decision tree would have grow infinitely - which is contradictory to the finite nature of decision trees.

I can't think of a proof that such a computation cannot be represented by a decision tree, and it's possible that my definitions aren't quite correct. But intuitively, I think this is the way it works.

Of course, if I'm wrong, any explanation on why and how would be very welcome.