## Question

Question Description Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ’s undirected graph serialization: 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 #.

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

## Solution

**Result:** Accepted **Time:** 78 ms

Here should be some explanations.

```
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
unordered_map<int,UndirectedGraphNode *> mp;
queue<UndirectedGraphNode *> que;
UndirectedGraphNode * root = NULL;
if(node)
{
root = new UndirectedGraphNode(node->label);
que.push(node); que.push(root);
mp[root->label] = root;
}
while(!que.empty())
{
UndirectedGraphNode * old_node = que.front();que.pop();
UndirectedGraphNode * new_node = que.front();que.pop();
for(const auto & x:old_node->neighbors)
if(mp.count(x->label))
new_node->neighbors.push_back(mp[x->label]);
else
{
UndirectedGraphNode * tmp = new UndirectedGraphNode(x->label);
new_node->neighbors.push_back(tmp);
mp[x->label] = tmp;
que.push(x);que.push(tmp);
}
}
return root;
}
};
```

**Complexity Analytics**

- Time Complexity:
- Space Complexity: