BFS vs DFS: Which Traversal Should You Use?

BFS and DFS are the two ways to visit every node of a graph, and almost every graph algorithm is one of them with something extra bolted on. Choosing correctly between them is usually the first decision in a graph problem, and getting it wrong makes the rest of the solution much harder than it needs to be.

In this article we will learn how the two traversals actually differ, why BFS finds the shortest path in an unweighted graph and DFS does not, how their memory usage compares on wide versus deep graphs, and a decision rule you can apply in an interview.

What is breadth-first search?

What is depth-first search?

How do they differ in practice?

Consider this graph, with A as the start:

BFS visits: A, then B and C, then D, E and F, then G. Level by level.

DFS visits: A, B, D, G, backtrack, E, backtrack, C, F. One branch to the bottom, then the next.

Both reach G. BFS reaches it at distance 3 via A → B → D → G, and that is genuinely the shortest. DFS also happens to reach it via that route here, but only by luck of the ordering — reorder the adjacency lists and DFS could find a longer path first and accept it.

The implementations

from collections import deque
 
def bfs(graph, start):
    visited = {start}
    queue = deque([start])
    order = []
 
    while queue:
        node = queue.popleft()      # FIFO — this is what makes it breadth-first
        order.append(node)
        for neighbour in graph[node]:
            if neighbour not in visited:
                visited.add(neighbour)   # mark on enqueue, not on dequeue
                queue.append(neighbour)
 
    return order

def dfs(graph, start, visited=None, order=None):
    if visited is None:
        visited, order = set(), []
 
    visited.add(start)
    order.append(start)
 
    for neighbour in graph[start]:
        if neighbour not in visited:
            dfs(graph, neighbour, visited, order)
 
    return order

The only structural difference is the queue versus the stack. Swap popleft() for pop() in the BFS code and you have an iterative DFS.

Which one should you use?

The decision rule

Ask one question: does the problem care about distance?

• "Shortest", "minimum number of steps", "fewest moves", "level", "nearest" → BFS
• "Does a path exist", "all paths", "connected components", "detect a cycle", "topological order" → DFS

Two practical caveats. On a graph with weighted edges, neither is correct for shortest paths — you need Dijkstra. And if the graph might be very deep, recursive DFS can exhaust the call stack; convert it to an explicit stack.

Common mistakes

• Using DFS for a shortest-path question. The most common graph error in interviews.
• Marking visited at the wrong moment in BFS. See the callout above.
• Forgetting the visited set entirely. On a cyclic graph, both traversals loop forever.
• Recursing on a deep graph. Python defaults to a recursion limit around 1,000.
• Assuming BFS works for weighted graphs. It gives fewest edges, not lowest cost.

Conclusion

BFS and DFS have identical O(V + E) time complexity and differ only in the order they visit nodes: BFS in expanding rings from the source using a queue, DFS one branch at a time using a stack.

That ordering difference is decisive. BFS guarantees shortest paths in unweighted graphs and uses memory proportional to the widest level. DFS gives no path-length guarantee, uses memory proportional to the deepest path, and is the more natural fit for cycle detection, topological sorting and exhaustive path enumeration.

If distance matters, use BFS. Otherwise DFS is usually simpler.