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Problem solution · Python

Merge Two Binary Trees

Merge Two Binary Trees: 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
16 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 Merge Two Binary Trees, 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

  • 16 lines of Python from the credited upstream file 617.py.
  • The implementation keeps its working state in language-native values and containers.
  • 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 codeMerge Two Binary Trees · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def mergeTrees(      self,      root1: TreeNode | None,      root2: TreeNode | None,  ) -> TreeNode | None:    if not root1 and not root2:      return None    val = (root1.val if root1 else 0) + (root2.val if root2 else 0)    root = TreeNode(val)    root.left = self.mergeTrees(root1.left if root1 else None,                                root2.left if root2 else None)    root.right = self.mergeTrees(root1.right if root1 else None,                                 root2.right if root2 else None)    return root

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