133. Clone Graph

• Difficulty: Medium

• Topics: Depth-first Search, Breadth-first Search, Graph

• Similar Questions:

  • Copy List with Random Pointer

Problem:

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.

Solutions:

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> neighbors;

    Node() {}

    Node(int _val, vector<Node*> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
};
*/
class Solution {
public:
    Node* cloneGraph(Node* node) {
        unordered_map<Node*, Node*> oldToNew;
        return helper(node, oldToNew);
    }

    Node* helper(Node* node, unordered_map<Node*, Node*>& oldToNew) {
        if (node == NULL)   return NULL;
        if (oldToNew.count(node) > 0)   return oldToNew[node];
        Node* newNode = new Node(node->val, {});
        oldToNew[node] = newNode;
        for (auto& neighbor : node->neighbors) {
            newNode->neighbors.push_back(helper(neighbor, oldToNew));
        }

        return newNode;
    }
};