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.

Solution

DFS

Time complexity : O(V+E)
Space complexity : O(V+E)

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> neighbors;
    
    Node() {
        val = 0;
        neighbors = vector<Node*>();
    }
    
    Node(int _val) {
        val = _val;
        neighbors = vector<Node*>();
    }
    
    Node(int _val, vector<Node*> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
};
*/

class Solution {
public:
    Node* cloneGraph(Node* node) {
        if (nullptr == node) return node;
        if (0 < visited.count(node)) return visited[node];
        
        Node* cur = new Node(node->val);
        visited[node] = cur;
        for (Node* neighbor: node->neighbors) {
            cur->neighbors.push_back(cloneGraph(neighbor));
        }
        return cur;
    }
private:
    unordered_map<Node*, Node*> visited;
};

BFS

Time complexity : O(V+E)
Space complexity : O(V+E)

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> neighbors;
    
    Node() {
        val = 0;
        neighbors = vector<Node*>();
    }
    
    Node(int _val) {
        val = _val;
        neighbors = vector<Node*>();
    }
    
    Node(int _val, vector<Node*> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
};
*/

class Solution {
public:
    Node* cloneGraph(Node* node) {
        if (nullptr == node) return node;
        
        queue<Node*> q;
        q.push(node);
        Node* root = new Node(node->val);
        visited[node] = root;
        while (!q.empty()) {
            Node* cur = q.front(); q.pop();
            for (Node* neighbor: cur->neighbors) {
                if (0 == visited.count(neighbor)) {
                    visited[neighbor] = new Node(neighbor->val);
                    q.push(neighbor);
                }
                visited[cur]->neighbors.push_back(visited[neighbor]);
            }
        }
        
        return root;
    }
private:
    unordered_map<Node*, Node*> visited;
};

不管是DFS還是BFS，都建一個hash table存放原node相對應的新node，藉以了解新舊對應關係。

DFS使用recursive的方式寫，應該很好理解。
但因recursive的方式，若考量到stack空間，可改用BFS，iterative的方式寫。
BFS，iterative，就是用一個queue來存放下一階段要處理的neighbors。

剩下的新增node、連線，做法都差不多。