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In this post, you will find the solution for the **Clone Graph** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Clone Graph– LeetCode Problem

Clone Graph– LeetCode Problem

**Problem:**

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.

Clone Graph– LeetCode Solutions

Clone Graph Solution in C++:

class Solution { public: Node* cloneGraph(Node* node) { if (!node) return nullptr; if (map.count(node)) return map[node]; Node* newNode = new Node(node->val); map[node] = newNode; for (Node* neighbor : node->neighbors) newNode->neighbors.push_back(cloneGraph(neighbor)); return newNode; } private: unordered_map<Node*, Node*> map; };

Clone Graph Solution in Java:

class Solution { public Node cloneGraph(Node node) { if (node == null) return null; Queue<Node> q = new LinkedList<>(Arrays.asList(node)); Map<Node, Node> map = new HashMap<>(); map.put(node, new Node(node.val)); while (!q.isEmpty()) { Node n = q.poll(); for (Node neighbor : n.neighbors) { if (!map.containsKey(neighbor)) { map.put(neighbor, new Node(neighbor.val)); q.offer(neighbor); } map.get(n).neighbors.add(map.get(neighbor)); } } return map.get(node); } }

Clone Graph Solution in Python:

class Solution: def cloneGraph(self, node: 'Node') -> 'Node': if not node: return None if node in self.map: return self.map[node] newNode = Node(node.val, []) self.map[node] = newNode for neighbor in node.neighbors: self.map[node].neighbors.append(self.cloneGraph(neighbor)) return newNode map = {}