---
title: '133. Clone Graph'
description: Given a reference of a node in a connected undirected graph
icon: dot
topics:
  - { name: "Hash Table", slug: "hash-table" }
  - { name: "Depth-First Search", slug: "depth-first-search" }
  - { name: "Breadth-First Search", slug: "breadth-first-search" }
  - { name: "Graph Theory", slug: "graph" }
issue: "https://github.com/prdlk/leetcode/issues/131"
sidebar:
  label: 'Clone Graph'
  badge: 'Medium'
---

### 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.

### 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.

## Solution

```py
"""
# 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 []
"""

from typing import Optional


class Solution:
    def cloneGraph(self, node: Optional["Node"]) -> Optional["Node"]:
        oldToNew = {}

        def dfs(node):
            if node in oldToNew:
                return oldToNew[node]

            copy = Node(node.val)
            oldToNew[node] = copy

            for n in node.neighbors:
                copy.neighbors.append(dfs(n))
            return copy

        return dfs(node) if node else None
```