Clone Graph

Clone an undirected graph. Each node in the graph contains alabeland a list of itsneighbors.

How we serialize an undirected graph:

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 as1 . Connect node1 to node2 .
3. Third node is labeled as2 . Connect node2 to node2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

   1
  / \
 /   \
0 --- 2
     / \
     \_/

Example

return a deep copied graph.

Solution 1: DFS recursive.

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null) {
        return node;
    }    
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    map.put(node, new UndirectedGraphNode(node.label));
    dfs(node, map);
    return map.get(node);
}
public void dfs(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
    if(node == null) {
        return ;
    }
    for(UndirectedGraphNode n : node.neighbors) {
        if(!map.containsKey(n)) {
            UndirectedGraphNode newnode = new UndirectedGraphNode(n.label);
            map.put(n, newnode);
            dfs(n, map);
        }
        map.get(node).neighbors.add(map.get(n));
    }
}

Solution 2: DFS iterative.

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null) {
        return node;
    }    
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    dfs(node, map);
    return map.get(node);
}
public void dfs(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
    if(node == null) {
        return ;
    }
    Stack<UndirectedGraphNode> stack = new Stack<UndirectedGraphNode>();
    stack.push(node);
    map.put(node, new UndirectedGraphNode(node.label));
    while(!stack.isEmpty()) {
        UndirectedGraphNode tmp = stack.pop();
        for(UndirectedGraphNode n : tmp.neighbors) {
            if(!map.containsKey(n)) {
                UndirectedGraphNode newnode = new UndirectedGraphNode(n.label);
                map.put(n, newnode);
                stack.push(n);
            }    
            map.get(tmp).neighbors.add(map.get(n));
        }
    }
}

Solution 3: BFS.

public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
    if(node == null) {
        return node;
    }    
    HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
    bfs(node, map);
    return map.get(node);
}
public void bfs(UndirectedGraphNode node, HashMap<UndirectedGraphNode, UndirectedGraphNode> map) {
    if(node == null) {
        return ;
    }
    Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
    queue.offer(node);
    map.put(node, new UndirectedGraphNode(node.label));
    while(!queue.isEmpty()) {
        UndirectedGraphNode tmp = queue.poll();
        for(UndirectedGraphNode n : tmp.neighbors) {
            if(!map.containsKey(n)) {
                UndirectedGraphNode newnode = new UndirectedGraphNode(n.label);
                map.put(n, newnode);
                queue.offer(n);
            }    
            map.get(tmp).neighbors.add(map.get(n));
        }
    }
}