133. Clone Graph Leetcode Python

克隆无向图算法

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本文介绍了一种基于广度优先搜索(BFS)的无向图克隆算法，包括递归和迭代两种实现方式。该算法的时间复杂度为O(V+E)，其中V表示顶点数，E表示边数。文章提供了Python代码示例，并解释了如何通过维护节点字典来避免重复克隆。

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

There are two solutions for this problem both based on BFS

one iterative another recursive.

The time complexity if O(V+E) and space is the same.

we need to maintain the que.

1. Recursive:

# Definition for a undirected graph node
# class UndirectedGraphNode:
#     def __init__(self, x):
#         self.label = x
#         self.neighbors = []

class Solution:
    # @param node, a undirected graph node
    # @return a undirected graph node
    def __init__(self):
        self.dict={None:None}
    def cloneGraph(self, node):
        if node==None:
            return None
        head=UndirectedGraphNode(node.label)
        self.dict[node]=head
        for n in node.neighbors:
            if n in self.dict:
                head.neighbors.append(self.dict[n])
            else:
                neigh=self.cloneGraph(n)
                head.neighbors.append(neigh)
        return head

For iterative solution.

It looks more intuitive.

# Definition for a undirected graph node
# class UndirectedGraphNode:
#     def __init__(self, x):
#         self.label = x
#         self.neighbors = []

class Solution:
    # @param node, a undirected graph node
    # @return a undirected graph node
    def cloneGraph(self, node):
        if node==None:
            return node
        que=[node]
        head=UndirectedGraphNode(node.label)
        dict={node:head}
        while que:
            curr=que.pop(0)
            for n in curr.neighbors:
                if n in dict:
                    dict[curr].neighbors.append(dict[n])
                else:
                    que.append(n)
                    copy=UndirectedGraphNode(n.label)
                    dict[n]=copy
                    dict[curr].neighbors.append(copy)
        return head