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# Clone Graph LeetCode Solution – Queslers

## Problem – 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.

### Clone Graph LeetCode Solution in 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())   //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);
}
};
``````

### Clone Graph LeetCode Solution in Python

``````    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
``````

### Clone Graph LeetCode Solution in Java

``````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));
}
map.get(n.label).neighbors.add(map.get(neighbor.label)); //add neighbor to new created nodes
}
}

return newNode;
}
}
``````
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