Description
Given a root of an N-ary tree, return a deep copy (clone) of the tree.
Each node in the n-ary tree contains a val (int) and a list (List[Node]) of its children.
class Node {
public int val;
public List<Node> children;
}
Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Example 1:
Input: root = [1,null,3,2,4,null,5,6]
Output: [1,null,3,2,4,null,5,6]
Example 2:
Input: 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]
Output: [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]
Constraints:
- The depth of the n-ary tree is less than or equal to
1000.
- The total number of nodes is between
[0, 104].
Follow up: Can your solution work for the
graph problem?
Solutions
Solution 1
PythonJavaC++Go
"""
# 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 is None:
return None
children = [self.cloneTree(child) for child in root.children]
return Node(root.val, children)(code-box)
/*
// 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;
}
ArrayList<Node> children = new ArrayList<>();
for (Node child : root.children) {
children.add(cloneTree(child));
}
return new Node(root.val, children);
}
}(code-box)
/*
// 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) {
return root;
}
vector<Node*> children;
for (Node* child : root->children) {
children.emplace_back(cloneTree(child));
}
return new Node(root->val, children);
}
};(code-box)
/**
* Definition for a Node.
* type Node struct {
* Val int
* Children []*Node
* }
*/
func cloneTree(root *Node) *Node {
if root == nil {
return nil
}
children := []*Node{}
for _, child := range root.Children {
children = append(children, cloneTree(child))
}
return &Node{root.Val, children}
}(code-box)
