BFS和DFS解决LeetCode133. Clone Graph

133 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 #.

First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
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     / \     \_/

题目虽然讲了很多内容，包括OJ中序列化的图的表示法，但是题目所要求的复制一个图，实际上只是考察了图的遍历。根据给定的一个UndirectedGraphNode* node，由该点出发，遍历整个图，并以此创建一个新的和原来一样的无向图即可。
图的遍历可以用DFS和BFS实现。这题唯一需要注意的地方在于，在创建新图时，要确认新建结点的“时机”以及连接节点的方式，不要重复new同一个结点。例如，假设给定的图为：

    0 -- 1

使用DFS，先访问0，然后访问0的临接结点1，而在访问了1之后，要准确地将先前创建的新的0结点放入1的neighbors中，否则会导致图的结构错误。这里，我采用了map<int, UndirectedGraphNode*>来解决。既可以保证不重复创建结点，又能根据label准确找回原来创建的结点。

代码实现如下：

DFS:

// DFS versionclass Solution {public:    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {        if (!node) {            return NULL;        }        node_map[node->label] = new UndirectedGraphNode(node->label);        dfs(node);        return node_map[node->label];    }private:    map<int, UndirectedGraphNode*> node_map;    void dfs(UndirectedGraphNode* &node) {        for (int i = 0; i < node->neighbors.size(); i++) {            UndirectedGraphNode* neighbor = node->neighbors[i];            if (node_map[neighbor->label] == NULL) {                node_map[neighbor->label] = new UndirectedGraphNode(neighbor->label);                dfs(neighbor);            }            node_map[node->label]->neighbors.push_back(node_map[neighbor->label]);        }    }};

BFS:

// BFS version:class Solution {public:    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {    if (!node) {        return NULL;    }    map<int, UndirectedGraphNode*> node_map;    node_map[node->label] = new UndirectedGraphNode(node->label);    queue<UndirectedGraphNode*> que;    que.push(node);    while (!que.empty()) {        UndirectedGraphNode* temp = que.front();        que.pop();        for (int i = 0; i < temp->neighbors.size(); i++) {            UndirectedGraphNode* neighbor = temp->neighbors[i];            if (node_map[neighbor->label] == NULL) {                node_map[neighbor->label] = new UndirectedGraphNode(neighbor->label);                que.push(neighbor);            }            node_map[temp->label]->neighbors.push_back(node_map[neighbor->label]);        }    }    return node_map[node->label];    }};