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Problem solution · Python

Encode N-ary Tree to Binary Tree

Encode N-ary Tree to Binary Tree: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
34 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

Sliding window or two pointers

For Encode N-ary Tree to Binary Tree, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 34 lines of Python from the credited upstream file 431-2.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeEncode N-ary Tree to Binary Tree · PythonPython
Use this to learn the idea, then write your own version.
class Codec:  # Encodes an n-ary tree to a binary tree.  def encode(self, root: 'Node') -> TreeNode | None:    if not root:      return None     rootTreeNode = TreeNode(root.val)    if root.children:      rootTreeNode.left = self.encode(root.children[0])     # The parent for the rest of the children    currTreeNode = rootTreeNode.left     # Encode the rest of the children    for i in range(1, len(root.children)):      currTreeNode.right = self.encode(root.children[i])      currTreeNode = currTreeNode.right     return rootTreeNode   # Decodes your binary tree to an n-ary tree.  def decode(self, root: TreeNode | None) -> 'Node':    if not root:      return None     rootNode = Node(root.val, [])    currTreeNode = root.left     while currTreeNode:      rootNode.children.append(self.decode(currTreeNode))      currTreeNode = currTreeNode.right     return rootNode

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