# 133. Clone Graph

Given a reference of a node in a [**connected**](https://en.wikipedia.org/wiki/Connectivity_\(graph_theory\)#Connected_graph) undirected graph.

Return a [**deep copy**](https://en.wikipedia.org/wiki/Object_copying#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:**

![](https://assets.leetcode.com/uploads/2019/11/04/133_clone_graph_question.png)

<pre><code><strong>Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
</strong><strong>Output: [[2,4],[1,3],[2,4],[1,3]]
</strong><strong>Explanation: There are 4 nodes in the graph.
</strong>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).
</code></pre>

***My Solutions:***

```
class Solution {
    public Node cloneGraph(Node node) {
        if (node == null) return null;
        
        Node copy = new Node(node.val);
        Node[] visited = new Node[101];
        dfs(node, copy, visited);
        return copy;
    }
    
    private void dfs(Node node, Node copy, Node[] visited) {
        // store the current node at it's val index which will tell us that this node is now visited
        visited[copy.val] = copy; 
        
        // 遍历每个neighbors
        for (Node n : node.neighbors) {
            
            // this node has not been visited
            if (visited[n.val] == null) { 
                Node newNode = new Node(n.val);
                copy.neighbors.add(newNode);
                dfs(n, newNode, visited);
            } 
            // if this node has been visited, retrieve node from visited array 
            // and add it as the adjacent node of the copied node
            else {  
                // THIS IS THE POINT WHY WE USED NODE[] INSTEAD OF BOOLEAN[] ARRAY
                copy.neighbors.add(visited[n.val]); 
            }
        }
    }
}
```
