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.

#

,

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.

解法
使用一个map记录节点的值和节点指针。使用dfs深度优先搜索图，每次创建节点的时候就在map里面增加一条记录。
（注意：需要在一开始的时候就增加map记录，否则会出现死循环的问题，例如

｛0，0，0｝

这个图，如果等到找到节点的时候再增加记录则出错。）

错误代码

class Solution {
    map<int,  UndirectedGraphNode *> dict;
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == nullptr)
            return node;
        UndirectedGraphNode *root =new  UndirectedGraphNode(node->label);
        for(int i=0;i<node->neighbors.size();i++) {
            if(dict.find(node->neighbors[i]->label)==dict.end())
                 dict[node->neighbors[i]->label] = cloneGraph(node->neighbors[i]); //在这才记录可能死循环
            root->neighbors.push_back(dict[node->neighbors[i]->label]);
        }
        return root;
    }
};

正确代码

class Solution {
    map<int,  UndirectedGraphNode *> dict;
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == nullptr)
            return node;
        UndirectedGraphNode *root =new  UndirectedGraphNode(node->label);
        dict[root->label] = root; // 在这里记录数据
        for(int i=0;i<node->neighbors.size();i++) {
            if(dict.find(node->neighbors[i]->label)==dict.end())
               cloneGraph(node->neighbors[i]);
            root->neighbors.push_back(dict[node->neighbors[i]->label]);
        }
        return root;
    }
};