NEW TO LEET CODE,

@tozy.zard

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LeetLockinseen in 6 videos →leetlockin.com ↗

Why it worked

The video addresses a common pain point for aspiring software engineers (LeetCode problems) and offers a clear, concise solution with code. The creator's bio link promotes a relevant product (AI tutor for LeetCode), creating a strong incentive for viewers to engage further.

Summary

The video explains how to invert a binary tree using recursion, a common problem on LeetCode. It demonstrates the Python code for swapping the left and right children of each node and recursively calling the function on the subtrees.

Structure

1Introduce the problem: Invert Binary Tree.
2Explain the base case for recursion (empty tree).
3Swap the left and right children of the current node.
4Recursively call the function on the left and right subtrees.
5Return the modified root.

Product placement

LeetCode (website): It is the platform where the problem is presented and solved. It acts as a substrate for the video's content. Removing it would change the context of the problem being solved. LeetLockin (website): The creator's bio link leads to this page, which is an AI Tutor for LeetCode Patterns & Interview Prep. It acts as the subject of the promotion. Removing it would remove the primary call to action and the product being promoted.

Original caption

Invert binary tree #python #leetcode #sweintern #swe #softwareengineer

More from @tozy.zard

Their other posts in the index, biggest breakout first.

Call to action

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On-screen text

NEW TO LEET CODE, YOU NEED TO KNOW THIS PROBLEM. INVERT BINARY TREE. ROOT OF A BINARY TREE, INVERT THE TREE AND RETURN ITS ROOT. IT, AND I'LL TO DO THE CODE. GONNA DO THIS RECURSION, CASE IS GONNA BE IF THE ROOT IS NONE, GONNA RETURN. IS NONE, THEN WE RETURN. SO WE COULD TEMPORARY VARIABLE, IT EASIER. SO ROOT OUT RIGHT, WE WANT ROOT BE EQUAL TO ROOT OUT LEFT. SO WE'RE SWAPPING ROOT OUT RIGHT AND ROOT OUT LEFT. AND THEN BECAUSE WE JUST CHANGED WE ACTUALLY NEED BRUTE LEFT TO BE EQUAL TO TEMPORARY. SO TEMP. NOW WE ONLY SWAPPED THE BRUTE RIGHT AND BRUTE LEFT, BUT IN BRUTE RIGHT AND BRUTE LEFT, SWAPPED, AND WE INVERT THOSE. SO WE HAVE FUNCTION ON ITSELF.. SO THIS IS WHERE THE RECURSION PART COMES IN. BRUTE SELF INVERT DOT RIGHT, GONNA CALL IT ON ROOT DOT LEFT. INVERT TREE ROOT THEN IF WE JUST RETURN THE ROOT, IT'S GONNA BE AND WE'RE DONE.

Transcript

If you are new to LeetCode, you need to know this problem. Invert Binary Tree. So given the root of a binary tree, invert the tree and return its root. Think about it and I'll do the code. Okay, so we're gonna do this recursion, and our base case is gonna be if the root is None, then we're just gonna return. So if root is None, then we return. Now we have to swap them. So we could use a temporary variable, just to make it easier. So root.right, we want root.right to be equal to root.left. So we're swapping root.right and root.left. And then because we just changed root.right, we actually need root.left to be equal to temporary. So temp. Now we only swapped the root.right and root.left, but in root.right and root.left, swapped, and we invert those. So we have to call the function on itself. So this is where the recursion part comes in. We're gonna call self.invertTree on root.right, and we're gonna call it on root.left. So self.invertTree(root.right). Then if we just return the root, it's gonna be the inverted tree, and we're done.