## Saturday, November 30, 2019

### Problem Statement

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.

### Video Tutorial

You can find the detailed video tutorial here

### Thought Process

It's a basic graph problem that can be solved in 3 different ways: BFS using queue, DFS using recursion, DFS using stack(very similar to BFT using queue). The trick is using a map to keep a one-to-one mapping between the old nodes and the copied nodes, meanwhile, we can also use the map to avoid a cycle when performing BFS or DFS.

### Solutions

#### BFS using queue

Time Complexity: O(N)
Space Complexity: O(N) the extra queue and map

#### DFS using recursion

Time Complexity: O(N)
Space Complexity: O(N) the extra map

#### DFS using stack

Time Complexity: O(N)
Space Complexity: O(N) the extra queue and stack

The main test method

### References

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