133. Clone Graph

1. Description

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.

2. Example

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.

3. 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.

4. Solutions

Hash Table

n is the number of nodes
Time complexity: O($n^2$)
Space complexity: O(n)

class Solution {
public:
    Node *cloneGraph(Node *node) {
        if (node == nullptr) {
            return nullptr;
        }

        unordered_map<Node *, Node *> visited;
        visited[node] = new Node(node->val);
        queue<Node *> nodes{{node}};
        while (!nodes.empty()) {
            auto node = nodes.front();
            nodes.pop();

            visited[node]->neighbors.reserve(node->neighbors.size());
            for (const auto &neighbor : node->neighbors) {
                if (visited.find(neighbor) == visited.end()) {
                    visited[neighbor] = new Node(neighbor->val);
                    nodes.push(neighbor);
                }

                visited[node]->neighbors.emplace_back(visited[neighbor]);
            }
        }

        return visited[node];
    }
};