Graph Depth-First Search
DFS explores one path as deeply as possible before backtracking. It can be implemented recursively or with an explicit stack.
Start with this mental model
DFS explores one path as deeply as possible before backtracking. It can be implemented recursively or with an explicit stack.
How it works, without unnecessary jargon
Mark a node visited before exploring neighbors to avoid cycles.
Across a graph represented by adjacency lists, traversal is O(V + E).
Read the algorithm as a sequence of decisions
Define directed/undirected and weighted/unweighted assumptions.
Choose the graph representation.
Track algorithm state and handle disconnected/cyclic cases explicitly.
Read a tiny input by hand and write down the state after every important operation. For Graph Depth-First Search, the goal is to make the invariant visible rather than to memorize code.
function dfs(graph, start) {
const seen = new Set();
function visit(node) {
if (seen.has(node)) return;
seen.add(node);
for (const next of graph[node] ?? []) visit(next);
}
visit(start);
return seen;
}Do not memorize the syntax. Follow the state that changes on each line and connect it to the invariant described above.
O(V+E)
Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.
Good fit
- Connectivity
- Cycle/SCC building blocks
- Backtracking-style graph exploration
What usually goes wrong
- Forgetting disconnected components.
- Mixing weighted and unweighted shortest-path algorithms.
- Not protecting against revisits/cycles.
A professional answer shape
State the visited-set requirement and O(V + E). Compare with BFS when shortest unweighted paths are needed.
What problem does Graph Depth-First Search solve?
DFS explores one path as deeply as possible before backtracking. It can be implemented recursively or with an explicit stack.
What should I mention about complexity?
Its typical cost here is O(V+E), but a complete answer should connect that cost to the representation, input constraints and the dominant operations.
What is the most common mistake?
Forgetting disconnected components.