BFS——克隆图，发现直接copy会出现某些环路的边会丢失，还是要先copy节点，再copy边

137. 克隆图

中文

English

克隆一张无向图. 无向图的每个节点包含一个 label 和一个列表 neighbors. 保证每个节点的 label 互不相同.

你的程序需要返回一个经过深度拷贝的新图. 新图和原图具有同样的结构, 并且对新图的任何改动不会对原图造成任何影响.

样例

样例1

输入:
{1,2,4#2,1,4#4,1,2}
输出: 
{1,2,4#2,1,4#4,1,2}
解释:
1------2  
 \     |  
  \    |  
   \   |  
    \  |  
      4   

说明

关于无向图的表示: http://www.lintcode.com/help/graph/

注意事项

你需要返回与给定节点具有相同 label 的那个节点.

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

""" class UndirectedGraphNode:      def __init__(self, x):          self.label = x          self.neighbors = [] """ from collections import deque     class Solution:     """     @param node: A undirected graph node     @return: A undirected graph node     """     def cloneGraph(self, node):         # write your code here         if not node:             return None                   def copy_graph_nodes(node):             nodes = {}             seen = {node}             q = deque([node])             while q:                 n = q.popleft()                 new_node = UndirectedGraphNode(n.label)                 nodes[n] = new_node                 for neighbor in n.neighbors:                     if neighbor not in seen:                         seen.add(neighbor)                         q.append(neighbor)                           return nodes                   def build_dir(node, nodes):             q = deque([node])             seen = {node}             while q:                 n = q.popleft()                 for neighbor in n.neighbors:                     nodes[n].neighbors.append(nodes[neighbor])                     if neighbor not in seen:                         seen.add(neighbor)                         q.append(neighbor)                                                             nodes = copy_graph_nodes(node)         build_dir(node, nodes)                   return nodes[node]