133. 克隆图

原题

https://leetcode.cn/problems/clone-graph/description/

思路

深度优先搜索

复杂度

时间：O(n)
空间：O(n)

Python代码

"""
# 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']:
        # 字典存储节点和克隆的节点
        d = dict()

        def dfs(node):
            if not node:
                return None 
            if node in d:
                return d[node]
            clone = Node(node.val, [])
            # 更新字典
            d[node] = clone 
            for neighbor in node.neighbors:
                clone.neighbors.append(dfs(neighbor))
            return clone
        
        return dfs(node)
       

Go代码

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Neighbors []*Node
 * }
 */

func cloneGraph(node *Node) *Node {
	// 字典存储节点和克隆的节点
	d := make(map[*Node]*Node)
	var dfs func(*Node) *Node
	// 返回节点对应的克隆节点
	dfs = func(node *Node) *Node {
		if node == nil {
			return nil
		}
		if d[node] != nil {
			return d[node]
		}
		clone := &Node{Val: node.Val}
		// 更新字典
		d[node] = clone
		for _, neighbor := range node.Neighbors {
			clone.Neighbors = append(clone.Neighbors, dfs(neighbor))
		}
		return clone
	}
	return dfs(node)
}