Clone Graph LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Clone Graph in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the Problem – Clone Graph– LeetCode Problem

Clone Graph– LeetCode Problem

Problem:

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 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:

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.

Constraints:

• The number of nodes in the graph is in the range [0, 100].
• 1 <= Node.val <= 100
• Node.val is unique for each node.
• There are 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.

Clone Graph– LeetCode Solutions

Clone Graph Solution in C++:

class Solution {
 public:
  Node* cloneGraph(Node* node) {
    if (!node)
      return nullptr;
    if (map.count(node))
      return map[node];

    Node* newNode = new Node(node->val);
    map[node] = newNode;

    for (Node* neighbor : node->neighbors)
      newNode->neighbors.push_back(cloneGraph(neighbor));

    return newNode;
  }

 private:
  unordered_map<Node*, Node*> map;
};

Clone Graph Solution in Java:

class Solution {
  public Node cloneGraph(Node node) {
    if (node == null)
      return null;

    Queue<Node> q = new LinkedList<>(Arrays.asList(node));
    Map<Node, Node> map = new HashMap<>();
    map.put(node, new Node(node.val));

    while (!q.isEmpty()) {
      Node n = q.poll();
      for (Node neighbor : n.neighbors) {
        if (!map.containsKey(neighbor)) {
          map.put(neighbor, new Node(neighbor.val));
          q.offer(neighbor);
        }
        map.get(n).neighbors.add(map.get(neighbor));
      }
    }

    return map.get(node);
  }
}

Clone Graph Solution in Python:

class Solution:
  def cloneGraph(self, node: 'Node') -> 'Node':
    if not node:
      return None
    if node in self.map:
      return self.map[node]

    newNode = Node(node.val, [])
    self.map[node] = newNode

    for neighbor in node.neighbors:
      self.map[node].neighbors.append(self.cloneGraph(neighbor))

    return newNode

  map = {}