Given a reference of a node in a connected undirected graph, return a deep copy (clone) of the graph. Each node in the graph contains a val (int) and a list (List[Node]) of its neighbors.

Example:

Input:
{"$id":"1","neighbors":[{"$id":"2","neighbors":[{"$ref":"1"},{"$id":"3","neighbors":[{"$ref":"2"},{"$id":"4","neighbors":[{"$ref":"3"},{"$ref":"1"}],"val":4}],"val":3}],"val":2},{"$ref":"4"}],"val":1}

Explanation:
Node 1's value is 1, and it has two neighbors: Node 2 and 4.
Node 2's value is 2, and it has two neighbors: Node 1 and 3.
Node 3's value is 3, and it has two neighbors: Node 2 and 4.
Node 4's value is 4, and it has two neighbors: Node 1 and 3.

Note:

1. The number of nodes will be between 1 and 100.
2. The undirected graph is a simple graph, which means no repeated edges and no self-loops in the graph.
3. Since the graph is undirected, if node  p  has node  q  as neighbor, then node  q  must have node  p  as neighbor too.
4. You must return the copy of the given node as a reference to the cloned graph.

这道无向图的复制问题和之前的 Copy List with Random Pointer 有些类似，那道题的难点是如何处理每个结点的随机指针，这道题目的难点在于如何处理每个结点的 neighbors，由于在深度拷贝每一个结点后，还要将其所有 neighbors 放到一个 vector 中，而如何避免重复拷贝呢？这道题好就好在所有结点值不同，所以我们可以使用 HashMap 来对应原图中的结点和新生成的克隆图中的结点。对于图的遍历的两大基本方法是深度优先搜索 DFS 和广度优先搜索 BFS，这里我们先使用深度优先搜索DFS来解答此题，在递归函数中，首先判空，然后再看当前的结点是否已经被克隆过了，若在 HashMap 中存在，则直接返回其映射结点。否则就克隆当前结点，并在 HashMap 中建立映射，然后遍历当前结点的所有 neihbor 结点，调用递归函数并且加到克隆结点的 neighbors 数组中即可，代码如下：

解法一：

class Solution {
public:
    Node* cloneGraph(Node* node) {
        unordered_map<Node*, Node*> m;
        return helper(node, m);
    }
    Node* helper(Node* node, unordered_map<Node*, Node*>& m) {
        if (!node) return NULL;
        if (m.count(node)) return m[node];
        Node *clone = new Node(node->val);
        m[node] = clone;
        for (Node *neighbor : node->neighbors) {
            clone->neighbors.push_back(helper(neighbor, m));
        }
        return clone;
    }
};

我们也可以使用 BFS 来遍历图，使用队列 queue 进行辅助，还是需要一个 HashMap 来建立原图结点和克隆结点之间的映射。先克隆当前结点，然后建立映射，并加入 queue 中，进行 while 循环。在循环中，取出队首结点，遍历其所有 neighbor 结点，若不在 HashMap 中，我们根据 neigbor 结点值克隆一个新 neighbor 结点，建立映射，并且排入 queue 中。然后将 neighbor 结点在 HashMap 中的映射结点加入到克隆结点的 neighbors 数组中即可，参见代码如下：

解法二：

class Solution {
public:
    Node* cloneGraph(Node* node) {
        if (!node) return NULL;
        unordered_map<Node*, Node*> m;
        queue<Node*> q{{node}};
        Node *clone = new Node(node->val);
        m[node] = clone;
        while (!q.empty()) {
            Node *t = q.front(); q.pop();
            for (Node *neighbor : t->neighbors) {
                if (!m.count(neighbor)) {
                    m[neighbor] = new Node(neighbor->val);
                    q.push(neighbor);
                }
                m[t]->neighbors.push_back(m[neighbor]);
            }
        }
        return clone;
    }
};

类似题目：

Copy List with Random Pointer

参考资料：

https://leetcode.com/problems/clone-graph/

https://leetcode.com/problems/clone-graph/discuss/42313/C%2B%2B-BFSDFS

https://leetcode.com/problems/clone-graph/discuss/42309/Depth-First-Simple-Java-Solution

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