LeetCode 133. Clone Graph

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 #.
　　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
     / \
     \_/

简单来说，就是让我们实现克隆一个无向图，克隆的意思是再创造一个一模一样的无向图，需要另外申请内存空间，不能直接原封不动的把输入直接作为结果返回 (/ □ \)

题目描述中使用了邻接链表来表示这个无向图，要克隆这个无向图，我们必定要用到图的搜索，由起点开始，根据原图的边信息来访问原图中的每个结点，同时拷贝这个结点及相关边的信息，即可完成图的克隆。对于图的搜索我们可以使用BFS或DFS，使用BFS要创建一个队列，使用DFS可以使用递归或创建一个栈，这里使用DFS递归求解本题。

需要注意的是，根据边的信息访问到原图中的某一结点，该结点在此之前有可能已经被访问过了，比如对题干例子的第一趟无回退深搜访问路径 0 ——> 1 ——> 2 ——> 2，就已经重复访问了结点2，所以我们需要对访问过的结点进行标记，为了节省空间这里使用HashMap来进行标记。代码如下

/**
 * Definition for undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     List<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 * };
 */
public class Solution {
    
    private HashMap<Integer, UndirectedGraphNode> labelToNode = new HashMap<>();
    
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        
        return clone(node);
    }
    
    private UndirectedGraphNode clone(UndirectedGraphNode node)
    {
        if(node == null)
            return null;
        if (labelToNode.containsKey(node.label))    // 已被访问且拷贝过的结点
            return labelToNode.get(node.label);
        UndirectedGraphNode copy = new UndirectedGraphNode(node.label); // 没有被访问过的新结点
        labelToNode.put(copy.label, copy);
        for(UndirectedGraphNode nbr : node.neighbors)
            copy.neighbors.add(clone(nbr));
        return copy;
    }
}