## The challenge

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.

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.

**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;
}
}
```

Code language: Java (java)

## 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) {
map.get(node).neighbors.add(map.containsKey(n)
? map.get(n)
: recurse(n, map));
}
}
return map.get(node);
}
}
```

Code language: Java (java)