LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal (Algorithm Explained)

Coolll easiest explanation on internet

Thank You So Much for this wonderful video.......🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

ur explanations are not clear and understandable you should be first clear urself and then teach others

Thank you Nick!

Why code in this video get stack over flow?

Pretty stupid explanation tbh

His attitude is kind of the worse, It's extremely hard to get through this because of it. I can feel condescension permeating this video.

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Thanks Nick. You are awesome! Constructing Binary Tree from PostOrder and Inorder is very similar to this problem. In PostOrder, the sequence is "left, right, center", which means we need to loop PostOrder array from the end. Recursive inputs to root.left and root.right will also differ accordingly (but very similar to this video).

Not sure how he has so much followers. The solutions are either taken from solution or reporduced from memory without any explanations.

Thanks for all ur work!

complex

Yeah, this was a doozy

Thank you ! really helpful!

Your solutions always explain the intuition behind the solution.

Good explanation of preorder, inorder and postorder

Beauty Nick !!

Nick, can you share your 2ms submission? Thanks

Anyone think question is hard? I even saw this video and I still don’t get it.

Sorry but this video sucks. This seems like some crappy memorization over understanding approach.

Your solution is always very concise. I was coding it and got all the index mangled up. Thanks for the video!

You can convert the in order array to a hash map storing {inorder val : index} to avoid looping through the in order array each time

Nice Video, gr8 explaination

you get my stuff. Watch and subscribe so that I can grow my thing. What is he referring to

thanks nick

not sure why people dislike these videos?

But in other sites like gfg they are passing preStart as prestart+1 in both root->left and root->right and still giving correct output. Can you explain why?

good explaination!

Review

Kevin >> Nick

Thanks for simplifying the problem!! Can you tell time complexity and space complexity of this?

Is there a way to improve look up for the inIndex?

You can save the value and index lookup for the inorder in a hashmap to avoid the for loop in your recursive stack.

could you please zoom out a little further, please? Who wants to see what's being typed on the screen anyways

(inIndex - inStart) is basically leftSubTreeCount which is here only 1 as there is only 1 node i.e. 9 on the left side of 3. So, that's why preorder start value will be preStart + leftSubtreeCount + 1 here.

Awesome explanation. Thanks

so u knoe how to reverse a linked list, but do u knoe how to design a large scale software system

Man! It does not look like you in this video, you must have drive some vodka...

I think the `preStart>preorder.length-1` check is not needed at all. For the left downward calls, eventually inEnd will reach 0 & it will base case out because inEnd decreases & inStart does not. For the right downward calls, inStart will base case out because it increases & inEnd does not. Also FYI it sort of feels like you're "just" reading us the solutions shown on Leetcode during some parts of the video (breaking eye contact & reading the other person's answer as a reference/running into various errors/erratic explanations). Overall its a great video, this is just some constructive criticism.

It took me 1 hour to do this problem, and you just did it in 11 minutes DAM

Just subscribed cause you said you can grow as you get more subscribers LOL We will help your channel grow, you help us learn data structures and algorithms. I appreciate you teaching us these concepts.

Great video!!! I took a while to understand the concept. Can you tell what is the time and space complexity for this approach?

Most of this explanation just bounced over my head! Hey Nick, can you please explain the algorithm a bit more clearly, before getting into the code. Because end of the day, logic and derivation of the solution is what matters right, and coding will be easy after that! Thanks for the video though.

good explanation but the screen is far....i can't see obviously

In line 29, where does inIndex - inStart + 1 come from? In other words, how do you map inorder index to preorder index?

nailed it . thanks for the clear explaination

Thanks for the video, it helped me. Do you know what would be the approach if the array would have repeated elements? (or the work around)

it is hard to understand the formula for the prestart of the right nodes. My solution was passing prestart as reference and increment its by 1 for every-time a node is created. Thank you for your video.

I find that the main time cost in this solution is from the for loop search part, you can reduce the time cost by using a hashmap for inorder element search.