【leetcode】133. Clone Graph

题目如下：

Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.

OJ's undirected graph serialization (so you can understand error output):

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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. 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
         / \
         \_/

Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer. You don't need to understand the serialization to solve the problem.

解题思路：深拷贝node。题目本身不难，由于所有节点的lable都是唯一的，因此需要保持已经创建过节点的lable，避免出现重复创建。另外节点存在self-cycle，所以遍历过的路径也需要保存。

代码如下：

# 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 None
        root = UndirectedGraphNode(node.label)
        queue = [(node,root)]
        dic = {}
        dic[root.label] = root
        dic_visit = {}
        while len(queue) > 0:
            n,r = queue.pop(0)
            if n.label in dic_visit:
                continue
            for i in n.neighbors:
                if i.label not in dic:
                    i_node = UndirectedGraphNode(i.label)
                    dic[i.label] = i_node
                else:
                    i_node = dic[i.label]
                r.neighbors.append(i_node)
                queue.append((i, r.neighbors[-1]))
            dic_visit[n.label] = 1

        return root