[Leetcode] 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
         / \
         \_/

Analysis:

/**
 * Definition for undirected graph.
 * struct UndirectedGraphNode {
 *     int label;
 *     vector<UndirectedGraphNode *> neighbors;
 *     UndirectedGraphNode(int x) : label(x) {};
 * };
 */
class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if (node == NULL)
            return NULL;
            
        //use BFS to visit all nodes
        queue<UndirectedGraphNode*> q;
        //use hash table to marked as visited
        map<UndirectedGraphNode*, UndirectedGraphNode*> m;
        m[node] = new UndirectedGraphNode(node->label);
        q.push(node);
        while(!q.empty())
        {
            UndirectedGraphNode* t = q.front();
            q.pop();
            for(int i=0; i<t->neighbors.size(); i++)
            {
                UndirectedGraphNode* n = t->neighbors[i];
                //if not visited before, added to the hash table and queue
                if (m.find(n) == m.end())
                {
                    m[n] = new UndirectedGraphNode(n->label);
                    q.push(n);
                }
                
                m[t]->neighbors.push_back(m[n]);
            }
        }
        
        return m[node];
    }
};