Last updated on July 18th, 2024 at 04:17 am

Here, We see Clone Graph LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Breadth-First Search, Depth-First Search, Graph

Companies

Facebook, Google, Uber, Pocketgems

Level of Question

Medium

Clone Graph LeetCode Solution

Problem Statement

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.

1. Clone Graph LeetCode Solution C++

class Solution {
public:
    Node* dfs(Node* cur,unordered_map<Node*,Node*>& mp)
    {
        vector<Node*> neighbour;
        Node* clone=new Node(cur->val);
        mp[cur]=clone;
            for(auto it:cur->neighbors)
            {
                if(mp.find(it)!=mp.end())
                {
                    neighbour.push_back(mp[it]);
                }
                else
                    neighbour.push_back(dfs(it,mp));
            }
            clone->neighbors=neighbour;
            return clone;
    }
    Node* cloneGraph(Node* node) {
        unordered_map<Node*,Node*> mp;
        if(node==NULL)
            return NULL;
        if(node->neighbors.size()==0)
        {
            Node* clone= new Node(node->val);
            return clone; 
        }
        return dfs(node,mp);
    }
};

2. Clone Graph LeetCode Solution Java

class Solution {
    public Node cloneGraph(Node node) {
        if (node == null) {
            return null;
        }
        Map<Node, Node> visited = new HashMap<>();
        return cloneGraphHelper(node, visited);
    }
    
    private Node cloneGraphHelper(Node node, Map<Node, Node> visited) {
        Node copy = new Node(node.val);
        visited.put(node, copy);
        for (Node neighbor : node.neighbors) {
            if (visited.containsKey(neighbor)) {
                copy.neighbors.add(visited.get(neighbor));
            } else {
                Node neighborCopy = cloneGraphHelper(neighbor, visited);
                copy.neighbors.add(neighborCopy);
            }
        }
        return copy;
    }
}

3. Clone Graph LeetCode Solution JavaScript

var cloneGraph = function(node) {
    let start = node; 
    if (start === null) return null;
    const vertexMap = new Map(); 
    const queue = [start]
    vertexMap.set(start, new Node(start.val));
    while (queue.length > 0) {
        const currentVertex = queue.shift();
        for (const neighbor of currentVertex.neighbors) {
            if (!vertexMap.has(neighbor)) {
                vertexMap.set(neighbor, new Node(neighbor.val))
                queue.push(neighbor); 
            }
            vertexMap.get(currentVertex).neighbors.push(vertexMap.get(neighbor)); 
        }
    }
    return vertexMap.get(start); 
};

4. Clone Graph Solution Python

class Solution(object):
    def cloneGraph(self, node):
        if not node:
            return None
        cloned = {}
        stack = [node]
        cloned[node] = Node(node.val)
        while stack:
            curr = stack.pop()
            for neighbor in curr.neighbors:
                if neighbor not in cloned:
                    cloned[neighbor] = Node(neighbor.val)
                    stack.append(neighbor)
                cloned[curr].neighbors.append(cloned[neighbor])
        return cloned[node]