Clone Graph

Description

Clone Graph

Given a reference of a node in a connected undirected graph.
Return a deep copy (clone) of the graph.
Each node in the graph contains a value (int) and a list (List[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
}

Test case format:
For simplicity, each node’s value is the same as the node’s index (1-indexed). For example, the first node with val == 1, the second node with val == 2, and so on. The graph is represented in the test case using an adjacency list.

An adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)‘s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)‘s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)‘s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)‘s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Example 2:

Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:
Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.

Constraints:

• The number of nodes in the graph is in the range [0, 100].
• 1 <= Node.val <= 100
• Node.val is unique for each node.
• There are no repeated edges and no self-loops in the graph.
• The Graph is connected and all nodes can be visited starting from the given node.

Primary Approach: DFS with HashMap ($O(V+E)$ Time, $O(V)$ Space)

Intuition

Cloning a graph requires duplicating every node and re-establishing all directed/undirected edges between the newly cloned nodes:

1. Use a HashMap (originalNode -> clonedNode) to keep track of already created clones. This serves as both our visited set and a mapping lookup.
2. Traverse the graph using Depth-First Search (DFS) starting from node.
3. If the current node is already in the map, return its saved clone (prevents infinite recursion on cycles).
4. Otherwise, instantiate a new Node(node.val), register it in the HashMap, and recursively clone all of its neighbors to populate its neighbors list.

import java.util.HashMap;
import java.util.Map;
 
class Solution {
    private Map<Node, Node> visited = new HashMap<>();
 
    public Node cloneGraph(Node node) {
        if (node == null) return null;
 
        // Return cloned instance if already visited to break cycles
        if (visited.containsKey(node)) {
            return visited.get(node);
        }
 
        // Create deep copy for current node
        Node cloneNode = new Node(node.val);
        visited.put(node, cloneNode);
 
        // Recursively clone all neighbors
        for (Node neighbor : node.neighbors) {
            cloneNode.neighbors.add(cloneGraph(neighbor));
        }
 
        return cloneNode;
    }
}
 

Complexity

• Time Complexity: $O(V+E)$ — Every vertex $V$ and edge $E$ in the graph is visited exactly once.
• Space Complexity: $O(V)$ — HashMap stores $V$ key-value pairs, and the recursion call stack takes up to $O(V)$ depth in a linear graph.

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Alternative Approach: BFS with HashMap ($O(V+E)$ Time, $O(V)$ Space)

Intuition

Perform a level-by-level traversal using an explicit Queue:

1. Store cloned nodes in a HashMap (originalNode -> clonedNode).
2. Push the starting node into a Queue and insert its clone into the HashMap.
3. While the queue is not empty, poll curr. For each neighbor in curr.neighbors:
   • If neighbor hasn’t been cloned yet, create its clone, put it in the map, and push neighbor to the queue.
   • Add visited.get(neighbor) to visited.get(curr).neighbors.

import java.util.ArrayDeque;
import java.util.HashMap;
import java.util.Map;
import java.util.Queue;
 
class Solution {
    public Node cloneGraph(Node node) {
        if (node == null) return null;
 
        Map<Node, Node> visited = new HashMap<>();
        Queue<Node> queue = new ArrayDeque<>();
 
        // Initialize root clone and BFS queue
        visited.put(node, new Node(node.val));
        queue.add(node);
 
        while (!queue.isEmpty()) {
            Node curr = queue.poll();
 
            for (Node neighbor : curr.neighbors) {
                // If neighbor hasn't been cloned yet, clone and queue it
                if (!visited.containsKey(neighbor)) {
                    visited.put(neighbor, new Node(neighbor.val));
                    queue.add(neighbor);
                }
                // Connect cloned current node to cloned neighbor
                visited.get(curr).neighbors.add(visited.get(neighbor));
            }
        }
 
        return visited.get(node);
    }
}
 

Complexity

• Time Complexity: $O(V+E)$ — Each node and edge is processed once during queue operations.
• Space Complexity: $O(V)$ — Queue holds at most $O(V)$ nodes, and HashMap stores $V$ cloned nodes.

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Key Interview Discussion Points

• Why HashMap is Necessary: Graph nodes can contain cycles (e.g., $1\leftrightarrow 2$). Without a HashMap mapping original nodes to cloned instances, simple traversal would result in infinite loops or duplicated node creations.
• Deep Copy Verification: Remind the interviewer that returning original node references within neighbor lists violates deep copying requirements; every newly linked neighbor must point strictly to newly allocated Node memory addresses.

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Easy Memory Rule

“HashMap stores original -> cloned $\to$ Traverse via DFS/BFS $\to$ If cloned return it, else create clone, map it, and recursively populate neighbors!”