Leetcode 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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. 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
         / \
         \_/

/**
 * Definition for undirected graph.
 * public class UndirectedGraphNode {
 *     public int label;
 *     public IList<UndirectedGraphNode> neighbors;
 *     public UndirectedGraphNode(int x) { label = x; neighbors = new List<UndirectedGraphNode>(); }
 * };
 */

public class Solution {
    public UndirectedGraphNode CloneGraph(UndirectedGraphNode node) {
        if (node == null) return null;
        
        var root = new UndirectedGraphNode(node.label);
        var visited = new Dictionary<int, UndirectedGraphNode>();
        visited[root.label] = root;
        
        DFS(node, root, visited);
        
        return root;
    }
    
    private void DFS(UndirectedGraphNode node, UndirectedGraphNode clone, Dictionary<int, UndirectedGraphNode> visited)
    {
        if (node == null) return;
        
        foreach (var n in node.neighbors)
        {
            // note: because graph can have cycle, we need to make sure to not creating the node multi-times
            if (visited.ContainsKey(n.label))
            {   
                clone.neighbors.Add(visited[n.label]);
            }
            else
            {
                var c = new UndirectedGraphNode(n.label);
                clone.neighbors.Add(c);
                visited[n.label] = c;
                DFS(n, c, visited);
            }
        }
    }
}