133. 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 value (int) and a list (List[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
}

Test case format:

For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with val == 1, the second node with val == 2, and so on. The graph is represented in the test case using an adjacency list.

An adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]
Output: [[2,4],[1,3],[2,4],[1,3]]
Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Example 2:

Input: adjList = [[]]
Output: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.

Solution:

DFS

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> neighbors;
    public Node() {
        val = 0;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val) {
        val = _val;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val, ArrayList<Node> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
}
*/

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

        Map<Node, Node> map = new HashMap<>();

        Node result = dfs(node, map);
        return result;
    }

    private Node dfs(Node node, Map<Node, Node> map){
        Node newNode = new Node(node.val);
        map.put(node, newNode);

        for (Node n : node.neighbors){
            if (!map.containsKey(n)){
                newNode.neighbors.add(dfs(n , map));
            }else{
                newNode.neighbors.add(map.get(n));
            }

        }
        return newNode;
    }


}
// TC: O(V + E)
// SC: O(V)

This is a connected graph, so we can use the following way, if not we have to add whole nodes first.

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> neighbors;
    public Node() {
        val = 0;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val) {
        val = _val;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val, ArrayList<Node> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
}
*/

class Solution {
    Map<Node,Node> map = new HashMap<Node, Node>();
    public Node cloneGraph(Node node) {
        if (node == null){
            return null;
        }

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

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

        map.put(node, clone);

        for (Node neighbor : node.neighbors){
            clone.neighbors.add(cloneGraph(neighbor));
        }
        return clone;
    }
}

// TC:O(n)
// sC:O(n)