133. Clone Graph(图的复制)

---

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 #.

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
         / 
         \_/

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.

DFS

/**
 * 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;
        if (mp.find(node->label) == mp.end()){ 
        mp[node->label] = new UndirectedGraphNode(node -> label);
        for (UndirectedGraphNode* neigh : node -> neighbors)
             mp[node->label] -> neighbors.push_back(cloneGraph(neigh));
        }
        return mp[node->label];
        
    }
private:
    unordered_map<int, UndirectedGraphNode*> mp;
};