**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

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.

```
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()) //already clone and stored in map
{
neighbour.push_back(mp[it]); //directly push back the clone node from map to neigh
}
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) //if only one node present no neighbors
{
Node* clone= new Node(node->val);
return clone;
}
return dfs(node,mp);
}
};
```

```
def cloneGraph(self, node: 'Node') -> 'Node':
if not node:
return
nodeCopy = Node(node.val, [])
dic = {node: nodeCopy}
queue = collections.deque([node])
while queue:
node = queue.popleft()
for neighbor in node.neighbors:
if neighbor not in dic: # neighbor is not visited
neighborCopy = Node(neighbor.val, [])
dic[neighbor] = neighborCopy
dic[node].neighbors.append(neighborCopy)
queue.append(neighbor)
else:
dic[node].neighbors.append(dic[neighbor])
return nodeCopy
```

```
public class Solution {
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) return null;
UndirectedGraphNode newNode = new UndirectedGraphNode(node.label); //new node for return
HashMap<Integer, UndirectedGraphNode> map = new HashMap(); //store visited nodes
map.put(newNode.label, newNode); //add first node to HashMap
LinkedList<UndirectedGraphNode> queue = new LinkedList(); //to store **original** nodes need to be visited
queue.add(node); //add first **original** node to queue
while (!queue.isEmpty()) { //if more nodes need to be visited
UndirectedGraphNode n = queue.pop(); //search first node in the queue
for (UndirectedGraphNode neighbor : n.neighbors) {
if (!map.containsKey(neighbor.label)) { //add to map and queue if this node hasn't been searched before
map.put(neighbor.label, new UndirectedGraphNode(neighbor.label));
queue.add(neighbor);
}
map.get(n.label).neighbors.add(map.get(neighbor.label)); //add neighbor to new created nodes
}
}
return newNode;
}
}
```

In our experience, we suggest you solve this Clone Graph LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Clone Graph LeetCode Solution

I hope this Clone Graph LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

**More Coding Solutions >>**