133. Clone Graph【M】【73】【图的遍历】【再来一遍】【vip】

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
/ \
\_/

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https://leetcode.com/discuss/57285/python-solutions-bfs-dfs-iteratively-dfs-recursively

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

class Solution(object):
def cloneGraph(self, node):

if not node:
return

res = {}
root = UndirectedGraphNode(node.label)
res[node] = root
n = node
q =

while q:
cur = q[0]
q = q[1:]

for i in cur.neighbors:
if i not in res:
q += i,
nb = UndirectedGraphNode(i.label)
res[i] = nb
res[cur].neighbors += nb,
else:
#原来写成了，这样一来，就是把没有复制的直接给放到新的里面去了。。
#res[cur].neighbors += i,
res[cur].neighbors += res[i],

return root#res[node.label]