133. Clone Graph (Medium)

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

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

class Solution {
    // record the visited node to avoid multi times access
    Map<Node, Node> visited = new HashMap<>();

    public Node cloneGraph(Node node) {
        if (node == null) {
            return null;
        }

        if (visited.containsKey(node)) {
            return visited.get(node);
        }

        Node newNode = new Node(node.val, new ArrayList<>());

        visited.put(node, newNode);

        if (node.neighbors == null) {
            return newNode;
        }

        for (Node n : node.neighbors) {
            // DFS technique to go through and clone all children nodes
            Node xn = cloneGraph(n);

            newNode.neighbors.add(xn);
        }

        return newNode;
    }
}

Incorrect Solutions

References