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 Listneighbors; }
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.
Solution: DFS + Hashtable



Time complexity: O(V+E)
Space complexity: O(V+E)
C++
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// Author: Huahua class Solution { public: Node* cloneGraph(Node* node) { if (!node) return nullptr; unordered_map<Node*, Node*> m; function<void(Node*)> dfs = [&](Node* u) { m[u] = new Node(u->val); for (Node* v : u->neighbors) { if (!m.count(v)) dfs(v); m[u]->neighbors.push_back(m[v]); } }; dfs(node); return m[node]; } }; |
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