### Question

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 `#`

.

- First node is labeled as
`0`

. Connect node`0`

to both nodes`1`

and`2`

. - Second node is labeled as
`1`

. Connect node`1`

to node`2`

. - 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 / \ \_/

### Stats

Adjusted Difficulty | 3 |

Time to use | thought for a while |

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

### Analysis

**This is a difficult coding question, although the idea is simple**.

### Solution

Two solution: **BFS (recommended) and DFS**. This is a good analysis article.

Let’s analyze this further by using the below example:

Assume that the starting point of the graph is A. First, you make a copy of node A (A2), and found that A has only one neighbor B. You make a copy of B (B2) and connects A2->B2 by pushing B2 as A2′s neighbor. Next, you find that B has A as neighbor, which you have already made a copy of. Here, we have to be careful not to make a copy of A again, but to connect B2->A2 by pushing A2 as B2′s neighbor. But, how do we know if a node has already been copied?

**Basic idea is to use HashMap to store the already-copied nodes**.

My first attempt is DFS by making use of a ‘visited’ Set to mark which node I have copied and which is not. This is a nice idea and it solved the problem neatly.

But after reading this article, I realize that **‘visited’ is not needed for BFS solution**!

**The trick is, whenever I do a ‘HashMap.put(curNode, newNode)’, I push ‘curNode’ to queue**. This very well replaces the functionality of the ‘visited’ set. It also guarantees that when I pop a new element from the queue, **I CAN ALWAYS FIND ITS CORRESPONDING COPY** from the HashMap - always there.

### Code

**First, my DFS code**

```
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) return null;
UndirectedGraphNode newNode = new UndirectedGraphNode(node.label);
HashMap<Integer, UndirectedGraphNode> map =
new HashMap<Integer, UndirectedGraphNode>();
map.put(node.label, newNode);
copy(node, newNode, map, new HashSet<Integer>());
return newNode;
}
private void copy(UndirectedGraphNode orin, UndirectedGraphNode cp,
HashMap<Integer, UndirectedGraphNode> map,
HashSet<Integer> visited) {
if (visited.contains(orin.label))
return;
else
visited.add(orin.label);
for (UndirectedGraphNode n : orin.neighbors) {
if (map.containsKey(n.label)) {
cp.neighbors.add(map.get(n.label));
} else {
UndirectedGraphNode newNode = new UndirectedGraphNode(n.label);
map.put(n.label, newNode);
cp.neighbors.add(map.get(n.label));
}
}
// do DFS recursively
for (int i = 0; i < orin.neighbors.size(); i++) {
copy(orin.neighbors.get(i), cp.neighbors.get(i), map, visited);
}
}
```

**Second, my BFS code without using ‘visited’ HashSet**

```
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) return null;
HashMap<UndirectedGraphNode, UndirectedGraphNode> map
= new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
map.put(node, new UndirectedGraphNode(node.label));
LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
queue.add(node);
// queue is guaranteed to always have non-traversed nodes
while ( !queue.isEmpty() ) {
UndirectedGraphNode orin = queue.remove();
UndirectedGraphNode cp = map.get(orin);
for (UndirectedGraphNode n : orin.neighbors) {
if ( !map.containsKey(n) ) {
map.put(n, new UndirectedGraphNode(n.label));
queue.add(n);
}
UndirectedGraphNode newNode = map.get(n);
cp.neighbors.add(newNode);
}
}
return map.get(node);
}
```