### 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

- Leetcode official solution (download pdf)
- Code Ganker CSDN

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