# 133. Clone Graph (Medium)

https://leetcode.com/problems/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 val (`int`) and a list (`List[Node]`) of its neighbors.

Example:

```Input:
{"\$id":"1","neighbors":[{"\$id":"2","neighbors":[{"\$ref":"1"},{"\$id":"3","neighbors":[{"\$ref":"2"},{"\$id":"4","neighbors":[{"\$ref":"3"},{"\$ref":"1"}],"val":4}],"val":3}],"val":2},{"\$ref":"4"}],"val":1}

Explanation:
Node 1's value is 1, and it has two neighbors: Node 2 and 4.
Node 2's value is 2, and it has two neighbors: Node 1 and 3.
Node 3's value is 3, and it has two neighbors: Node 2 and 4.
Node 4's value is 4, and it has two neighbors: Node 1 and 3.
```

Note:

1. The number of nodes will be between 1 and 100.
2. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
3. Since the graph is undirected, if node p has node q as neighbor, then node q must have node p as neighbor too.
4. You must return the copy of the given node as a reference to the cloned graph.

## Solutions

``````class Solution {
// record the visited node to avoid multi times access
Map<Node, Node> visited = new HashMap<>();

public Node cloneGraph(Node node) {
if (node == null) {
return null;
}

if (visited.containsKey(node)) {
return visited.get(node);
}

Node newNode = new Node(node.val, new ArrayList<>());

visited.put(node, newNode);

if (node.neighbors == null) {
return newNode;
}

for (Node n : node.neighbors) {
// DFS technique to go through and clone all children nodes
Node xn = cloneGraph(n);

}

return newNode;
}
}
``````