算法描述：

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.

Nodes are labeled uniquely.

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.

解題思路：可以采用深度優先搜索，也可以采用廣度優先搜索。這裏用廣度優先搜索進行遍歷，用一個map標記訪問過的點，並用一個隊列輔助遍歷。

    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == nullptr) return nullptr;
        unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> map;
        queue<UndirectedGraphNode*> que;
        que.push(node);
        map[node] = new UndirectedGraphNode(node->label);
        while(!que.empty()){
            UndirectedGraphNode* t = que.front();
            que.pop();
            for(auto neighbor:t->neighbors){
                if(map.find(neighbor)==map.end()){
                    map[neighbor] = new UndirectedGraphNode(neighbor->label);
                    que.push(neighbor);
                }
                map[t]->neighbors.push_back(map[neighbor]);
            }
        }
        return map[node];
    }

LeetCode-133-Clone Graph