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.
- The number of nodes in the graph is in the range
[0, 100]. 1 <= Node.val <= 100Node.valis 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.
"""
# Definition for a Node.
class Node:
def __init__(self, val = 0, neighbors = None):
self.val = val
self.neighbors = neighbors if neighbors is not None else []
"""
from typing import Optional
class Solution:
def cloneGraph(self, node: Optional['Node']) -> Optional['Node']:
def dfs(node):
if not node:
return None
if node in visited:
return visited[node]
copy = Node(node.val)
visited[node] = copy
for neighbor in node.neighbors:
copy.neighbors.append(dfs(neighbor))
return copy
visited = {}
return dfs(node)The time complexity is O(N + M), where N is the number of nodes and M is the number of edges in the graph. This is because we visit each node exactly once and clone each edge exactly once.
The space complexity is O(N) for the recursion stack and the mapping from original nodes to their copies. In the worst case, the recursion stack could grow to the size of the number of nodes in the graph.