133. Clone Graph
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.
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
"""
# 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 []
"""
class Solution:
def cloneGraph(self, node: 'Node') -> 'Node':
if not node:
return node
m = {node: Node(node.val)}
deq = deque([node])
while deq:
n = deq.popleft()
for nei in n.neighbors:
if nei not in m:
deq.append(nei)
m[nei] = Node(nei.val)
m[n].neighbors.append(m[nei])
return m[node]