Here's an answer in the simply-laced case. Its proof, and generalization to non-simply-laced, are left to the reader.
Start with a simple root, and think of it as a labeling of the Dynkin diagram with a 1 there and 0s elsewhere.
Look for a vertex whose label is < 1/2 the sum of the surrounding labels. Increment that label. You've found a root!
Go back to (2), unless there is no such vertex anymore. You've found the highest root!
Take the union over all such games, and you get all the positive roots. Include their negatives, and you have all roots.
If you start with a non-Dynkin-diagram, the game doesn't terminate. This is part of a way to classify the Dynkin diagrams.
BTW at the highest root, most of the vertices have labels = 1/2 the surrounding sum. If you put in a new vertex, connected to those vertices with > 1/2 the sum, you get the affine Dynkin diagram.