# 133. Clone Graph

## Problem

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 #.

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
/ \
\_/
```

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.

## Code

```
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> neighbors;
public Node() {}
public Node(int _val,List<Node> _neighbors) {
val = _val;
neighbors = _neighbors;
}
};
*/
class Solution {
public Node cloneGraph(Node node) {
if(node == null) return node;
HashMap<Node, Node> map = new HashMap<>();
// 两步: 先copy点, 再copy边
copyNode(node, map);
copyNei(node, map, new HashSet<Node>());
return map.get(node);
}
private void copyNode(Node node, HashMap<Node,Node> map){
if(map.containsKey(node)) return;
map.put(node, new Node(node.val, new ArrayList<>()));
for(Node nei : node.neighbors){
copyNode(nei, map);
}
}
private void copyNei(Node node, HashMap<Node, Node> map, HashSet<Node> visited){
if(visited.contains(node)) return;
visited.add(node);
for(Node nei : node.neighbors){
map.get(node).neighbors.add(map.get(nei));
copyNei(nei, map, visited);
}
}
}
```