For graph as follow:

The topological order can be:

<code>[0, 1, 2, 3, 4, 5]
[0, 2, 3, 1, 5, 4]
...
</code>

You can assume that there is at least one topological order in the graph.

Can you do it in both BFS and DFS?

Solution 1. BFS

1. first create a map which contains all the nodes and its indegrees

2. only add node with 0 indegree to the list and once the node is select, remove 1 indegree from all its neighbors

public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
    //BFS
    ArrayList<DirectedGraphNode> result = new ArrayList<>();
    if (graph == null || graph.size() <= 1) {
        return graph;
    }
    HashMap<DirectedGraphNode, Integer> inDegreeMap = new HashMap<>();
    for (DirectedGraphNode node : graph) {
        if (!inDegreeMap.containsKey(node)) {
            inDegreeMap.put(node, 0);
        }
        for (DirectedGraphNode neighbor : node.neighbors) {
            if (inDegreeMap.containsKey(neighbor)) {
                inDegreeMap.put(neighbor, inDegreeMap.get(neighbor) + 1);
            } else {
                inDegreeMap.put(neighbor, 1);
            }
        }
    }
    while (!inDegreeMap.isEmpty()) {
        for (DirectedGraphNode node : inDegreeMap.keySet()) {
            if (inDegreeMap.get(node) == 0) {
                result.add(node);
                inDegreeMap.remove(node);
                for (DirectedGraphNode neighbor : node.neighbors) {
                    inDegreeMap.put(neighbor, inDegreeMap.get(neighbor) - 1);
                }
                break;
            }
        }
    }
    return result;
}