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.

Submitted Code

"""
# 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']:
        if not node: return

        def dfs(curr_node):
            if curr_node in cloned_map:             # 이미 복사된 노드인지 확인
                return cloned_map[curr_node]
            copy = Node(curr_node.val)              # 현재 노드 복사
            cloned_map[curr_node] = copy
            for n in curr_node.neighbors:           # 원본 노드의 이웃들 조사
                copy.neighbors.append(dfs(n))
            return copy

        cloned_map = {}
        return dfs(node)

Runtime: 52 ms | Beats 43.09%
Memory: 19.88 MB | Beats 18.61%

이웃 노드를 단순히 그대로 복사해버리면 원본 노드의 주소를 다시 가리키게 되기 때문에 해시 테이블로 방문 여부를 확인했다.

Other Solutions

1st

class Solution(object):
    def cloneGraph(self, node):
        """
        :type node: Node
        :rtype: Node
        """
        if not node:
            return None
        
        cloned = {}
        stack = [node]
        cloned[node] = Node(node.val)
        
        while stack:
            curr = stack.pop()
            
            for neighbor in curr.neighbors:
                if neighbor not in cloned:
                    cloned[neighbor] = Node(neighbor.val)
                    stack.append(neighbor)
                
                cloned[curr].neighbors.append(cloned[neighbor])
        
        return cloned[node]

time complexity: 𝑂(𝑛+𝑒) ← n=number of nodes, e=number of edges
space complexity: 𝑂(𝑛)