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Complete guide to graphs — representation, BFS, DFS, shortest path algorithms, topological sort, and classic graph problems in Java.
What is a Graph?
A graph G = (V, E) consists of vertices (nodes) and edges (connections). Unlike trees, graphs can have cycles, multiple paths, and no root.
| Undirected Graph | Directed Graph (Digraph): |
| 1 --- 2 1 | 2 |
| 3 --- 4 3 | 4 |
BFS (Breadth-First Search) — O(V + E)
Example
public List<Integer> bfs(int start) {
List<Integer> order = new ArrayList<>();
boolean[] visited = new boolean[vertices];
Queue<Integer> queue = new LinkedList<>();
visited[start] = true;
queue.offer(start);
while (!queue.isEmpty()) {
int node = queue.poll();
order.add(node);
for (int[] neighbor : adjList.get(node)) {
if (!visited[neighbor[0]]) {
visited[neighbor[0]] = true;
queue.offer(neighbor[0]);
}
}
}
return order;
}
// Shortest path in unweighted graph
public int[] shortestPath(int source) {
int[] distance = new int[vertices];
Arrays.fill(distance, -1);
distance[source] = 0;
Queue<Integer> queue = new LinkedList<>();
queue.offer(source);
while (!queue.isEmpty()) {
int node = queue.poll();
for (int[] neighbor : adjList.get(node)) {
if (distance[neighbor[0]] == -1) {
distance[neighbor[0]] = distance[node] + 1;
queue.offer(neighbor[0]);
}
}
}
return distance;
}
DFS (Depth-First Search) — O(V + E)
Example
public List<Integer> dfs(int start) {
List<Integer> order = new ArrayList<>();
boolean[] visited = new boolean[vertices];
dfsHelper(start, visited, order);
return order;
}
private void dfsHelper(int node, boolean[] visited, List<Integer> order) {
visited[node] = true;
order.add(node);
for (int[] neighbor : adjList.get(node)) {
if (!visited[neighbor[0]]) {
dfsHelper(neighbor[0], visited, order);
}
}
}
// Iterative DFS
public List<Integer> dfsIterative(int start) {
List<Integer> order = new ArrayList<>();
boolean[] visited = new boolean[vertices];
Stack<Integer> stack = new Stack<>();
stack.push(start);
while (!stack.isEmpty()) {
int node = stack.pop();
if (visited[node]) continue;
visited[node] = true;
order.add(node);
for (int[] neighbor : adjList.get(node)) {
if (!visited[neighbor[0]]) stack.push(neighbor[0]);
}
}
return order;
}
Dijkstra's Shortest Path — O((V+E) log V)
Example
public int[] dijkstra(int source) {
int[] dist = new int[vertices];
Arrays.fill(dist, Integer.MAX_VALUE);
dist[source] = 0;
// Min-heap: [distance, vertex]
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
pq.offer(new int[]{0, source});
while (!pq.isEmpty()) {
int[] curr = pq.poll();
int d = curr[0], u = curr[1];
if (d > dist[u]) continue; // Outdated entry
for (int[] edge : adjList.get(u)) {
int v = edge[0], weight = edge[1];
if (dist[u] + weight < dist[v]) {
dist[v] = dist[u] + weight;
pq.offer(new int[]{dist[v], v});
}
}
}
return dist;
}
Topological Sort (DAG only) — O(V + E)
Example
// Kahn's Algorithm (BFS-based)
public List<Integer> topologicalSort() {
int[] indegree = new int[vertices];
for (int u = 0; u < vertices; u++)
for (int[] edge : adjList.get(u))
indegree[edge[0]]++;
Queue<Integer> queue = new LinkedList<>();
for (int i = 0; i < vertices; i++)
if (indegree[i] == 0) queue.offer(i);
List<Integer> order = new ArrayList<>();
while (!queue.isEmpty()) {
int node = queue.poll();
order.add(node);
for (int[] neighbor : adjList.get(node)) {
if (--indegree[neighbor[0]] == 0) queue.offer(neighbor[0]);
}
}
if (order.size() != vertices) throw new RuntimeException("Cycle detected!");
return order;
}
Cycle Detection
Example
// Undirected graph - using DFS
public boolean hasCycleUndirected() {
boolean[] visited = new boolean[vertices];
for (int i = 0; i < vertices; i++) {
if (!visited[i] && dfsCycleUndirected(i, -1, visited)) return true;
}
return false;
}
private boolean dfsCycleUndirected(int node, int parent, boolean[] visited) {
visited[node] = true;
for (int[] neighbor : adjList.get(node)) {
if (!visited[neighbor[0]]) {
if (dfsCycleUndirected(neighbor[0], node, visited)) return true;
} else if (neighbor[0] != parent) return true; // Back edge = cycle
}
return false;
}
// Directed graph - using colors (WHITE, GRAY, BLACK)
public boolean hasCycleDirected() {
int[] color = new int[vertices]; // 0=white, 1=gray, 2=black
for (int i = 0; i < vertices; i++) {
if (color[i] == 0 && dfsCycleDirected(i, color)) return true;
}
return false;
}
private boolean dfsCycleDirected(int node, int[] color) {
color[node] = 1; // Gray (being processed)
for (int[] neighbor : adjList.get(node)) {
if (color[neighbor[0]] == 1) return true; // Back edge to gray = cycle
if (color[neighbor[0]] == 0 && dfsCycleDirected(neighbor[0], color)) return true;
}
color[node] = 2; // Black (done)
return false;
}
Interview Questions
Q1: When to use BFS vs DFS?
Answer: BFS: shortest path in unweighted graphs, level-order processing, finding nearest node. DFS: topological sort, cycle detection, path finding, connected components, maze solving. BFS uses O(V) queue space; DFS uses O(V) stack space.
Q2: What is the difference between Dijkstra and BFS for shortest path?
Answer: BFS finds shortest path in unweighted graphs (all edges weight 1). Dijkstra handles weighted graphs with non-negative weights. BFS uses a regular queue; Dijkstra uses a priority queue.
Q3: How do you detect a cycle in a directed graph?
Answer: DFS with three states: WHITE (unvisited), GRAY (in current DFS path), BLACK (fully processed). A back edge to a GRAY node indicates a cycle. Alternatively, use Kahn's algorithm — if topological sort doesn't include all vertices, there's a cycle.
Q4: What is topological sorting and when is it used?
Answer: Linear ordering of vertices in a DAG such that for every edge u→v, u appears before v. Used for: task scheduling, build systems, course prerequisites, dependency resolution. Only possible for DAGs.
Q5: How do you find connected components?
Answer: Run DFS/BFS from each unvisited node. Each traversal discovers one connected component. For directed graphs, use Kosaraju's or Tarjan's algorithm to find strongly connected components.
Exam Focus
Revise definitions, diagrams, examples, and short-answer points for Graph Data Structure in Java.
Interview Use
Prepare one clear explanation, one practical example, and one common mistake for this Java Master Course topic.
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