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Problem solution · Python

All Elements in Two Binary Search Trees

All Elements in Two Binary Search Trees: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Sliding window or two pointers
Source
Kamyu LeetCode Solutions
Length
45 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 All Elements in Two Binary Search 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

  • 45 lines of Python from the credited upstream file all-elements-in-two-binary-search-trees.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 Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeAll Elements in Two Binary Search Trees · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(h) # Definition for a binary tree node.class TreeNode(object):    def __init__(self, x):        self.val = x        self.left = None        self.right = None  class Solution(object):    def getAllElements(self, root1, root2):        """        :type root1: TreeNode        :type root2: TreeNode        :rtype: List[int]        """        def inorder_gen(root):            result, stack = [], [(root, False)]            while stack:                root, is_visited = stack.pop()                if root is None:                    continue                if is_visited:                    yield root.val                else:                    stack.append((root.right, False))                    stack.append((root, True))                    stack.append((root.left, False))            yield None                result = []        left_gen, right_gen = inorder_gen(root1), inorder_gen(root2)        left, right = next(left_gen), next(right_gen)        while left is not None or right is not None:            if right is None or (left is not None and left < right):                result.append(left)                left = next(left_gen)            else:                result.append(right)                right = next(right_gen)        return result

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