133. Clone Graph

Given a reference of a node in a connected undirected 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 []
"""

from typing import Optional


class Solution:
    def cloneGraph(self, node: Optional["Node"]) -> Optional["Node"]:
        oldToNew = {}

        def dfs(node):
            if node in oldToNew:
                return oldToNew[node]

            copy = Node(node.val)
            oldToNew[node] = copy

            for n in node.neighbors:
                copy.neighbors.append(dfs(n))
            return copy

        return dfs(node) if node else None