Leetcode solution 133: Clone Graph

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.

Problem link

 

Video Tutorial

You can find the detailed video tutorial here

• Youtube
• B站

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