Clone Graph

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此题的答案非常简洁巧妙，看似用的是BFS，实则为DFS。

使用一个Map来记录克隆的旧新节点关系，然后设计一个函数，参数为node，表明这个函数要克隆node节点及其邻居关系。如果邻居就在Map里，直接返回，因为这是一个已经克隆过的邻居，否则，递归地克隆其邻居。

/** * Definition for undirected graph. * struct UndirectedGraphNode { *     int label; *     vector<UndirectedGraphNode *> neighbors; *     UndirectedGraphNode(int x) : label(x) {}; * }; */class Solution {public:    map<UndirectedGraphNode *, UndirectedGraphNode *> myMap;    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        return cloneNode(node);    }    UndirectedGraphNode *cloneNode(UndirectedGraphNode *node) {        if (node == NULL)            return NULL;        if (myMap.find(node) != myMap.end())            return myMap[node];        UndirectedGraphNode *t = new UndirectedGraphNode(node->label);        myMap[node] = t;        for (int i = 0; i < (node->neighbors).size(); ++i) {            t->neighbors.push_back(cloneNode((node->neighbors)[i]));        }        return t;    }};

http://oj.leetcode.com/problems/clone-graph/

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