# Topological Sorting

#### Topological Sorting

Given an directed graph, a topological order of the graph nodes is defined as follow:

• For each directed edge `A -> B` in graph, A must before B in the order list.
• The first node in the order can be any node in the graph with no nodes direct to it.

Find any topological order for the given graph.

Example

For graph as follow:

The topological order can be:

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

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

Challenge

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) {