LeetCode #133 - Clone Graph

题目描述：

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
         / \
         \_/

在图中每一个节点都有个数不确定的邻居，而且每个节点都有不同的标记。这种题目显然是用广度优先遍历，对每个邻居都进行遍历，没有遍历过的需要新建节点，为了避免重复遍历就要利用节点唯一的标记。

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        unordered_map<int, UndirectedGraphNode *> visited;
        vector<UndirectedGraphNode *> neighbors;
        return cloneGraph_recur(node, visited);
    }
    
    UndirectedGraphNode * cloneGraph_recur(UndirectedGraphNode *node, unordered_map<int,UndirectedGraphNode *>&visited)
    {
        if(node==NULL) return NULL;
        else if(visited.count(node->label)==0)
        {
            UndirectedGraphNode * newNode=new UndirectedGraphNode(node->label);
            visited[node->label]=newNode;
            for(int i=0;i<node->neighbors.size();i++)
                newNode->neighbors.push_back(cloneGraph_recur(node->neighbors[i],visited));
            return newNode;
        }
        else if(visited.count(node->label)>0) return visited[node->label];
    }
};