DSE Graph with Java Driver, how to walk through a graph (like a tree)

Question

I have a subgraph where I know how to come to the root vertex. But then I need to walk through it.

Concretely "walk through the subgraph" in my case means that I have to walk to all leafs of the subgraph (because I know that the subgraph is like a tree) and then go the path back and do some calculations between every vertex.

My question is, how to achieve that in the most performant way?

I can think about two solutions.

First, I walk through the graph with plenty of session.executeGraph("g.V().has('id','1')").one() statements to get all the single vertices and edges and do the calculations with them. But I think this way is very inefficient.

Or I work with the path object that I can get with

GraphNode node = session.executeGraph("g.V().has('id','1').repeat(outE().subgraph('sg').otherV()).cap('sg').path()").one();
Path path = node.asPath();

I am quite sure, the second solution is the preferred one but I have no clue how to use the path object to walk through the graph because the only thing I can see is a flat map of objects.

Update #1

Here is a picture of an example tree. The goal, I need the "combined value" for node A. The rules are quite simple. The nodes (except the root) has values. The edges has weightings. I have to sum all values regarding the weights. As long as a child has only one parent I can take the complete value. In case a child has multiple parents, I have to take the weighting into account. In the example tree, the combined value of B would be 100 + (500 * 50/60) + 1000 and the combined value of A would be combined value of B plus value of C (A == 2156.67). So, I need properties from the vertices and the edges for the calculation.

Update #2

So, here is my solution.

I've implemented an abstract Tree class that is doing the actual calculation (because I also have a mock implementation).

public abstract class Tree {
    // String == item id
    protected final Map<String, Item> items = new HashMap<>();
    private final String rootItemId;

    protected Tree(String rootItemId) {
        this.rootItemId = rootItemId;
    }

    public void accumulateExpenses() {
        accumulateExpenses(null, null);
    }

    private double accumulateExpenses(String itemId, String parentItemId) {
        final Item item = itemId == null ? items.get(rootItemId) : items.get(itemId);
        final double expense = item.getExpense();
        double childExpenses = 0;

        for (String childId : item.getChildIds()) {
            childExpenses += accumulateExpenses(childId, item.getId());
        }

        // calculate the percentage in case the item has multiple parents
        final double percentage = item.getPercentage(parentItemId);
        final double accumulatedExpenses = percentage * (expense + childExpenses);
        item.setAccumulatedExpense(accumulatedExpenses);

        return accumulatedExpenses;
    }
}

And I've implemented a GraphTree class that is responsible to fill the item map of the super class (abstract tree).

public class GraphTree extends Tree {
    public GraphTree(GraphNode graphNode, String rootNodeId) {
        super(rootNodeId);

        final GraphNode vertices = graphNode.get("vertices");
        final GraphNode edges = graphNode.get("edges");

        for (int i = 0; i < vertices.size(); i++) {
            final Vertex vertex = vertices.get(i).asVertex();
            final Item item = Item.fromVertex(vertex);
            super.items.put(item.getId(), item);
        }

        for (int i = 0; i < edges.size(); i++) {
            final Edge edge = edges.get(i).asEdge();
            final Relation relation = Relation.fromEdge(edge);
            super.items.get(relation.getParentId()).getRelations().add(relation);
        }
    }
}

For sake of completeness, here is also the Item class.

public class Item {
    private String id;
    private double accumulatedExpense;
    private final List<Relation> relations = new ArrayList<>();
    private final Map<String, Expense> expenses = new HashMap<>();

    public void setAccumulatedExpense(double accumulatedExpense) {
        this.accumulatedExpense = accumulatedExpense;
    }

    public double getPercentage(String parentId) {
        if (parentId == null) {
            return 1;
        }

        double totalWeight = 1;
        double weight = 1;

        for (Relation relation : relations) {
            if (Objects.equals(id, relation.getChildId())) {
                totalWeight += relation.getWeight();
                if (Objects.equals(parentId, relation.getParentId())) {
                    weight = relation.getWeight();
                }
            }
        }

        return weight / totalWeight;
    }

    public static Item fromVertex(Vertex vertex) {
        final Item item = new Item();
        item.setId(IdGenerator.generate(vertex));

        return item;
    }

    public List<String> getChildIds() {
        return relations.parallelStream()
                   .filter(relation -> Objects.equals(relation.getParentId(),id))
                   .map(Relation::getChildId)
                   .collect(Collectors.toList());
    }
}

To get the initial subgraph, I've used the following code.

    final String statement = String.format("g.V('%s').repeat(outE().subgraph('sg').otherV()).cap('sg')", rootNodeId);
    final GraphNode node = session.executeGraph(statement).one();

Answer 1

Even after reading the comments over and over again, I'm confused by the logic when I try to find a solution using a single query. Hence it's probably best to just tell you how to get a tree representation:

g.V().has('id','1').repeat(outE().as("e").inV()).emit(__.not(outE())).tree()

If you only need certain information (e.g. the value property of vertices and the weight property of edges), you can do this:

g.V().has('id','1').
  repeat(outE().as("e").inV()).emit(__.not(outE())).
  tree().by("value").by("weight")

And since vertex A doesn't seem to have a value property, you'll need to add a coalesce step:

g.V().has('id','1').
  repeat(outE().as("e").inV()).emit(__.not(outE())).
  tree().by(coalesce(values("value"), constant(0))).by("weight")

UPDATE

In case I need to play with the sample graph later again, here's the code to create it:

g = TinkerGraph.open().traversal()
g.addV().property(id, "A").as("a").
  addV().property(id, "B").property("value", 100).as("b").
  addV().property(id, "C").property("value", 200).as("c").
  addV().property(id, "D").property("value", 500).as("d").
  addV().property(id, "E").property("value", 1000).as("e").
  addV().property(id, "Z").property("value", 900).as("z").
  addE("link").from("a").to("b").property("weight", 80).
  addE("link").from("a").to("c").property("weight", 20).
  addE("link").from("b").to("d").property("weight", 50).
  addE("link").from("b").to("e").property("weight", 40).
  addE("link").from("z").to("d").property("weight", 10).iterate()