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.

Idea:

BFS Queue + HashMap +

• Time complexity: O(V+E) 点+边
• Space complexity: O(V+E)

Solution 1:

``````/**
* // Definition for a Node.
* function Node(val, neighbors) {
*    this.val = val === undefined ? 0 : val;
*    this.neighbors = neighbors === undefined ? [] : neighbors;
* };
*/
/**
* @param {Node} node
* @return {Node}
*/
var cloneGraph = function(node) {
if (node === null) return null;
let map = new Map();
map.set(node, new Node(node.val));
let q = []; // queue
q.push(node);
while (q.length !== 0) {
let cur = q.shift();
for (let n of cur.neighbors) {
if (!map.has(n)) {
map.set(n, new Node(n.val));
q.push(n);
}
map.get(cur).neighbors.push(map.get(n));
}
}
return map.get(node);
};``````

Solution 2:

``````var cloneGraph = function(node) {
if (node === null) return null;
let visited = new Map();
let map = new Map();
let q = []; // queue
q.push(node);
while (q.length !== 0) {
let cur = q.shift();
if (visited.has(cur)) continue;
visited.set(cur, true);
if (!map.has(cur)) map.set(cur, new Node(cur.val));
let temp = map.get(cur);
for (let n of cur.neighbors) {
if (!map.has(n)) map.set(n, new Node(n.val));
q.push(n);
temp.neighbors.push(map.get(n));
}
}
return map.get(node);
};``````