# Clone Graph LeetCode Solution | Easy Approach Share:

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 GraphSolutions

Time:O(∣V∣+∣E∣)
Space: O(|V| + |E|)O(∣V∣+∣E∣)

### C++

``````class Solution {
public:
Node* cloneGraph(Node* node) {
if (!node)
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];
}
};
``````

### 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);
}
}
}

return map.get(node);
}
}
``````

### 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]
``````

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