DFS - Recursive + Iterative

Easy Graphs BFS / DFS Basics

DFS - Recursive + Iterative is a easy graphs problem solved with the bfs / dfs basics pattern. The best approach, iterative dfs (explicit stack), runs in O(V + E) time and O(V + E) space. Below are 2 approaches in Java, from recursive dfs up.

Problem

Depth-first search follows one path as deep as it can, then backs up to the last branch point and tries the next option. Learn it recursively first, then with an explicit stack.

Examples

Example 1

Input: adj = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1]}, start = 0
Output: recursive order [0, 1, 3, 2]

Example 2

Input: same graph, iterative with a stack
Output: [0, 1, 3, 2] when neighbours are pushed in reverse

Constraints

V vertices, E edges, adjacency list input.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Recursive DFS	`O(V + E)`	`O(V)`
Iterative DFS (explicit stack)	`O(V + E)`	`O(V + E)`

1Recursive DFS

TimeO(V + E)

SpaceO(V)Recursion depth can reach V.

Mark the node, record it, then recurse into each unvisited neighbour. Returning from the recursion is the backtracking step.

dfs(u): seen[u] = true; record u.
For each v in adj[u]: if not seen, dfs(v).

Java

class Solution {
    public List<Integer> dfs(int V, List<List<Integer>> adj, int start) {
        List<Integer> order = new ArrayList<>();
        go(start, adj, new boolean[V], order);
        return order;
    }

    private void go(int u, List<List<Integer>> adj, boolean[] seen, List<Integer> order) {
        seen[u] = true;
        order.add(u);
        for (int v : adj.get(u)) if (!seen[v]) go(v, adj, seen, order);
    }
}

2Iterative DFS (explicit stack)

TimeO(V + E)

SpaceO(V + E)The stack can hold duplicate entries, up to E.

Push the start node. Pop a node; if already visited, skip it; otherwise visit it and push its neighbours. Push them in reverse to match the recursive order. Mark on pop, because a node can be pushed several times before it is visited.

stack = [start].
Pop u; if seen, continue; seen[u] = true; record u.
Push neighbours in reverse order.

Java

class Solution {
    public List<Integer> dfs(int V, List<List<Integer>> adj, int start) {
        List<Integer> order = new ArrayList<>();
        boolean[] seen = new boolean[V];
        Deque<Integer> st = new ArrayDeque<>();
        st.push(start);
        while (!st.isEmpty()) {
            int u = st.pop();
            if (seen[u]) continue;
            seen[u] = true;
            order.add(u);
            List<Integer> nb = adj.get(u);
            for (int i = nb.size() - 1; i >= 0; i--) if (!seen[nb.get(i)]) st.push(nb.get(i));
        }
        return order;
    }
}

Edge cases to test

Very deep graphs (recursion depth up to V; iterative avoids stack overflow)
Disconnected graph

Hints

Hint 1

DFS goes as deep as possible before backing up. The call stack is the stack; the iterative version makes it explicit.

FAQ

What is the best time complexity for DFS - Recursive + Iterative?

Iterative DFS (explicit stack) runs in O(V + E) time and O(V + E) extra space.

Which pattern does DFS - Recursive + Iterative use?

It is a graphs problem that uses the bfs / dfs basics pattern. Other problems with the same pattern: BFS - Normal + Level by Level.

Is there a brute force solution for DFS - Recursive + Iterative?

Yes. Recursive DFS takes O(V + E) time and O(V) space. Mark the node, record it, then recurse into each unvisited neighbour.

Which edge cases should I test for DFS - Recursive + Iterative?

Very deep graphs (recursion depth up to V; iterative avoids stack overflow); Disconnected graph.