[LeetCode]: 133: Clone Graph

题目：

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
         / 
         \_/

题目分析：

题目本身就是个图的遍历问题，其实只用看最上面的一句就OK了。下面整这么一篇就是为了说明Testcase，结果误导对了我好久。

与树的遍历不同之处就是图可能会出现"闭合的环"这样会造成无线循环，所以遍历的时候需要把已经遍历过的点去掉。

根据题目，图中每一个点的权值是不一样的，所以可以采用取巧的方法检测这个点是否已经加在了图中，即使用HashMap

思路一：深度优先搜索

    public static UndirectedGraphNode cloneDFS(UndirectedGraphNode node,HashMap<Integer, UndirectedGraphNode> validator){
        if(validator.containsKey(node.label)){
            return validator.get(node.label);
        }
        
        UndirectedGraphNode nodeTemp = new UndirectedGraphNode(node.label);
        validator.put(node.label, nodeTemp);
        
        for(int i =0;i<node.neighbors.size();i++){
            nodeTemp.neighbors.add(cloneDFS(node.neighbors.get(i),validator));
        }
        
        return nodeTemp;
    }
    public static UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if(node== null ||  node.equals(null)){
            return null;
        }
        
        HashMap<Integer, UndirectedGraphNode> validator = new HashMap<Integer, UndirectedGraphNode>();
        return cloneDFS(node, validator);
    }

思路二：广度优先搜索

XXX