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.

 ``````1 2 3 4 `````` ``````class Node { public int val; public List 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

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 `````` ``````class Node { public int val; public List neighbors; public Node() { val = 0; neighbors = new ArrayList(); } public Node(int _val) { val = _val; neighbors = new ArrayList(); } public Node(int _val, ArrayList _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.

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 `````` ``````class Solution { public Node cloneGraph(Node node) { return recurse(node, new HashMap()); } private static Node recurse(Node node, Map map) { if (node!=null) { map.put(node, new Node(node.val)); for(Node n: node.neighbors) { map.get(node).neighbors.add(map.containsKey(n) ? map.get(n) : recurse(n, map)); } } return map.get(node); } } ``````