# LeetCode 133. Clone Graph

## Description

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 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.
```

Example 4:

```Input: adjList = [,]
Output: [,]
```

Constraints:

• The number of nodes in the graph is in the range `[0, 100]`.
• `1 <= Node.val <= 100`
• `Node.val` is unique for each node.
• There are no repeated edges and no self-loops in the graph.
• The Graph is connected and all nodes can be visited starting from the given node.

## Explanation

We can use both depth-first search or breadth-first search to build a clone graph with help from a hashmap which maps the original nodes to the clone nodes.

## Python Solution

``````"""
# Definition for a Node.
class Node:
def __init__(self, val = 0, neighbors = None):
self.val = val
self.neighbors = neighbors if neighbors is not None else []
"""

class Solution:
def cloneGraph(self, node: 'Node') -> 'Node':
if not node:
return None
visited = {}
clone_root = self.helper(node, visited)

return clone_root

def helper(self, node, visited):
if not node:
return None

if node in visited:
return visited[node]

if not node.neighbors:
return Node(node.val)

clone = Node(node.val)

visited[node] = clone

for neighbor in node.neighbors:
clone_neighbor = self.helper(neighbor, visited)

clone.neighbors.append(clone_neighbor)

return clone``````
``````"""
# Definition for a Node.
class Node:
def __init__(self, val = 0, neighbors = None):
self.val = val
self.neighbors = neighbors if neighbors is not None else []
"""

class Solution:
def cloneGraph(self, node: 'Node') -> 'Node':
if not node:
return None

visited = {}

root = node
queue = []
queue.append(node)
visited[node] = Node(node.val)

while queue:

current_node = queue.pop(0)

for neighbor in current_node.neighbors:
if neighbor not in visited:
queue.append(neighbor)
visited[neighbor] = Node(neighbor.val)

visited[current_node].neighbors.append(visited[neighbor])

return visited[root]
``````
• Time Complexity: O(N).
• Space Complexity: O(N).