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70 lines
2.2 KiB
Plaintext
70 lines
2.2 KiB
Plaintext
---
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title: '133. Clone Graph'
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description: Given a reference of a node in a connected undirected graph
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icon: dot
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topics:
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- { name: "Hash Table", slug: "hash-table" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Breadth-First Search", slug: "breadth-first-search" }
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- { name: "Graph Theory", slug: "graph" }
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issue: "https://github.com/prdlk/leetcode/issues/131"
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sidebar:
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label: 'Clone Graph'
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badge: 'Medium'
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---
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### Example 1:
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- Input: `adjList = [[2,4],[1,3],[2,4],[1,3]]`
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- Output: `[[2,4],[1,3],[2,4],[1,3]]`
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- 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`).
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### Example 2:
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- Input: `adjList = [[]]`
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- Output: `[[]]`
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- 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.
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### Example 3:
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- Input: `adjList = []`
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- Output: `[]`
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- Explanation: This an empty graph, it does not have any nodes.
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### Constraints:
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- The number of nodes in the graph is in the range [0, 100].
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- `1 <= Node.val <= 100`
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- Node.val is unique for each node.
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- There are no repeated edges and no self-loops in the graph.
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- The Graph is connected and all nodes can be visited starting from the given node.
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## Solution
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```py
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"""
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# Definition for a Node.
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class Node:
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def __init__(self, val = 0, neighbors = None):
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self.val = val
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self.neighbors = neighbors if neighbors is not None else []
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"""
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from typing import Optional
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class Solution:
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def cloneGraph(self, node: Optional["Node"]) -> Optional["Node"]:
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oldToNew = {}
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def dfs(node):
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if node in oldToNew:
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return oldToNew[node]
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copy = Node(node.val)
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oldToNew[node] = copy
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for n in node.neighbors:
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copy.neighbors.append(dfs(n))
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return copy
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return dfs(node) if node else None
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```
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