LeetCode-clone_graph
克隆一个图， 用了两个遍历方法如下：
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.
Visually, the graph looks like the following:
1
/ \
/ \
0 --- 2
/ \
\_/
# _*_ coding: utf-8 _*_
# 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 cloneGraph1(self, node):
# BFS
if not node:
return
_queue = collections.deque([node])
record = {node.label: UndirectedGraphNode(node.label)}
while _queue:
item = _queue.popleft()
for nei in item.neighbors:
if nei.label not in record:
record[nei.label] = UndirectedGraphNode(nei.label)
_queue.append(nei)
record[item.label].neighbors.append(record[nei.label])
return record[node.label]
def cloneGraph(self, node):
# DFS
def dfs(node):
if node.label in self.record:
return self.record[node.label]
self.record[node.label] = UndirectedGraphNode(node.label)
for nei in node.neighbors:
self.record[node.label].neighbors.append(dfs(nei))
return self.record[node.label]
if not node:
return
self.record = {}
return dfs(node)