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Problem solution · Python

Clone N Ary Tree

Clone N Ary Tree: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
53 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Clone N Ary Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 53 lines of Python from the credited upstream file clone-n-ary-tree.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeClone N Ary Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(h) # Definition for a Node.class Node(object):    def __init__(self, val=None, children=None):        self.val = val        self.children = children if children is not None else []  class Solution(object):    def cloneTree(self, root):        """        :type root: Node        :rtype: Node        """        result = [None]        stk = [(1, (root, result))]        while stk:            step, params = stk.pop()            if step == 1:                node, ret = params                if not node:                    continue                ret[0] = Node(node.val)                for child in reversed(node.children):                    ret1 = [None]                    stk.append((2, (ret1, ret)))                    stk.append((1, (child, ret1)))            else:                ret1, ret = params                ret[0].children.append(ret1[0])        return result[0]  # Time:  O(n)# Space: O(h)class Solution2(object):    def cloneTree(self, root):        """        :type root: Node        :rtype: Node        """        def dfs(node):            if not node:                return None            copy = Node(node.val)            for child in node.children:                copy.children.append(dfs(child))            return copy                return dfs(root)

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