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.

链接：https://leetcode.com/problems/clone-graph/description/

题解：跟之前复制一个随机链表类似，直接用递归方法。

class Solution {
public:
    unordered_map<int,UndirectedGraphNode*> visited;
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(node == nullptr)
            return node;
        
        if(visited.find(node->label) != visited.end())
            return visited[node->label];

        UndirectedGraphNode *nwnode = new UndirectedGraphNode(node->label);
        visited[node->label] = nwnode;
        
        for(int i = 0;i < node->neighbors.size();i++){
            nwnode->neighbors.push_back(cloneGraph(node->neighbors[i]));    
        }
        
        return nwnode;
    }
};