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Problem solution · Python

Subtree of Another Tree

Subtree of Another Tree: 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
28 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 Subtree of Another 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

  • 28 lines of Python from the credited upstream file subtree-of-another-tree.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 Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSubtree of Another Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(m * n), m is the number of nodes of s, n is the number of nodes of t# Space: O(h), h is the height of s class Solution(object):    def isSubtree(self, s, t):        """        :type s: TreeNode        :type t: TreeNode        :rtype: bool        """        def isSame(x, y):            if not x and not y:                return True            if not x or not y:                return False            return x.val == y.val and \                   isSame(x.left, y.left) and \                   isSame(x.right, y.right)         def preOrderTraverse(s, t):            return s != None and \                   (isSame(s, t) or \                    preOrderTraverse(s.left, t) or \                    preOrderTraverse(s.right, t))         return preOrderTraverse(s, t)

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