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,2}.
The graph has a total of three nodes, and therefore contains three parts as separated by#.
First node is labeled as0. Connect node0to both nodes1and2.Second node is labeled as1. Connect node1to node2.Third node is labeled as2. Connect node2to node2(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 nodedef __init__(self):self.dict={None:None}def cloneGraph(self, node):if node==None:return Nonehead=UndirectedGraphNode(node.label)self.dict[node]=headfor 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 nodedef cloneGraph(self, node):if node==None:return nodeque=[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]=copydict[curr].neighbors.append(copy)return head
，用积极的拼搏迎接雨后的彩虹,相信自己
