In this post, we are going to solve the **Clone Graph** **Leetcode Solution** problem of **Leetcode**. This Leetcode problem is done in many programming languages like C++, Java, and Python.

**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.

Now, let’s see the leetcode solution of **Clone Graph** **Leetcode Solution**.

**Clone Graph Leetcode Solution in Python**

class Solution: def cloneGraph(self, node: 'Node') -> 'Node': if not node: return None q = deque([node]) map = {node: Node(node.val)} while q: u = q.popleft() for v in u.neighbors: if v not in map: map[v] = Node(v.val) q.append(v) map[u].neighbors.append(map[v]) return map[node]

**Clone Graph Leetcode Solution** **in CPP**

class Solution { public: Node* cloneGraph(Node* node) { if (node == nullptr) return nullptr; queue<Node*> q{{node}}; unordered_map<Node*, Node*> map{{node, new Node(node->val)}}; while (!q.empty()) { Node* u = q.front(); q.pop(); for (Node* v : u->neighbors) { if (!map.count(v)) { map[v] = new Node(v->val); q.push(v); } map[u]->neighbors.push_back(map[v]); } } return map[node]; } };

**Clone Graph Leetcode Solution in Java**

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

**Note:** This problem **Clone Graph** is generated by **Leetcode **but the solution is provided by **Chase2learn **This tutorial is only for Educational and Learning purpose.