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1490. 克隆 N 叉树

English Version

题目描述

给定一棵 N 叉树的根节点 root ，返回该树的深拷贝（克隆）。

N 叉树的每个节点都包含一个值（ int ）和子节点的列表（ List[Node] ）。

class Node {
    public int val;
    public List<Node> children;
}

N 叉树的输入序列用层序遍历表示，每组子节点用 null 分隔（见示例）。

进阶：你的答案可以适用于克隆图问题吗？

 

示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[1,null,3,2,4,null,5,6]

示例 2：

输入：root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出：[1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

 

提示：

  • 给定的 N 叉树的深度小于或等于 1000。
  • 节点的总个数在 [0, 10^4] 之间

解法

DFS。

Python3

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

class Solution:
    def cloneTree(self, root: 'Node') -> 'Node':
        if root:
            node = Node(val=root.val)
            node.children = [self.cloneTree(child) for child in root.children]
            return node

Java

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;


    public Node() {
        children = new ArrayList<Node>();
    }

    public Node(int _val) {
        val = _val;
        children = new ArrayList<Node>();
    }

    public Node(int _val,ArrayList<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public Node cloneTree(Node root) {
        if (root == null) {
            return null;
        }
        Node node = new Node(root.val);
        for (Node child : root.children) {
            node.children.add(cloneTree(child));
        }
        return node;
    }
}

C++

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    Node* cloneTree(Node* root) {
        if (root == nullptr) {
            return nullptr;
        }
        Node* node = new Node(root->val);
        vector<Node*> children;
        for (Node* node : root->children) {
            children.push_back(cloneTree(node));
        }
        node->children = children;
        return node;
    }
};

Go

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

func cloneTree(root *Node) *Node {
	if root == nil {
		return nil
	}
	node := &Node{Val: root.Val}
	for _, child := range root.Children {
		node.Children = append(node.Children, cloneTree(child))
	}
	return node
}

...