## The challenge#

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph. Effectively a graph copy.

Each node in the graph contains a val (`int`) and a list (`List[Node]`) of its neighbors.

``````class Node {
public int val;
public List<Node> neighbors;
}
``````

Test case format:

For simplicity sake, 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.

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. This will be a graph copy that we can use.

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.
```

Example 4:

```Input: adjList = [[2],[1]]
Output: [[2],[1]]
```

Constraints:

• `1 <= Node.val <= 100`
• `Node.val` is unique for each node.
• Number of Nodes will not exceed 100.
• There is 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.

## The Definition for a Node#

``````class Node {
public int val;
public List<Node> neighbors;

public Node() {
val = 0;
neighbors = new ArrayList<Node>();
}

public Node(int _val) {
val = _val;
neighbors = new ArrayList<Node>();
}

public Node(int _val, ArrayList<Node> _neighbors) {
val = _val;
neighbors = _neighbors;
}
}
``````

## The solution in Java#

We can solve this by means of a recursive traversal, as this is an undirected graph.

Start by visiting a node, move through it’s list of neighbours, call each neighbour recursively if they exist.

Implement a `map` to check which nodes have already been called.

``````class Solution {
public Node cloneGraph(Node node) {
return recurse(node, new HashMap<Node, Node>());
}

private static Node recurse(Node node, Map<Node, Node> map) {
if (node!=null) {
map.put(node, new Node(node.val));
for(Node n: node.neighbors) {