LOWEST COMMON ANCESTOR OF A BINARY TREE III [PYTHON] 

Man, honestly it is one of the underrated channel. Your explanation is crystal clear, easy to understand. I had difficulty in understanding complex problem's solutions, but your videos really helped me. Thanks a lot :)

The two pointer approach took me a while to understand but I think this rephrasing might help a few people. As mentioned in the video, P_copy takes 3 steps and Q_copy takes 2 steps. This results in a difference of 1 step which is not what we want. What we need to do is essentially make them do the same amount of work so we arrive at the same node. What we can do is combine them, so P_copy and Q_copy do the same amount of work So if we do the 'reset' as mentioned in the video: P_Copy does 3 + 2 steps and Q_Copy does 2 + 3 steps. This results in the difference being 0 so no matter where they are in the tree they will always end up at the same place!

The solution sounds like fast slow pointers technique and it is brilliant. Thanks a lot, ser.

Brilliant explanation! I got quite confused when I was browsing through the discussion sessions and saw these 5 lines of magic Lol... but your video just made it so clear!

You said the time complexity of naive solution is O(2N), but I just think it's O(N). Even the traverse is twice which is by p and also by q, total sum of them won't be greater than N.

So clear. Thank you very much

Thanks, bro, really good explanation, love it!

This channel is underrated! Great work. Keep the good work! :)

Awesome! Thank you!

This is such an excellent solution. I had to watch it twice though because it wasn't clear you would be swapping the pointers so I was wondering but once I figured that out I was like wow. One other thing is maybe calling them q_ptr and p_ptr would have been better and its odd that the examples require us to return values yet we return nodes

Basically same "ah-ha" trick as Leetcode 160 -- Intersection of Two Linked Lists

Thanks man, Now Understand similar question of finding intersection of two linked list.

Thank you, that explanation is very clear.

Great explanation. The two pointer solution is exactly similar to Intersection of two linked lists problem. In an interview, would we be expected to come up with such a solution or will the set solution suffice?

Everything is amazing in this video! With your explanation the second solution clicks pretty easily. But what's IQ one should have to be able to solve it at an interview, within 15-20 minutes and without seeing the solution before, it's probably around 160.

Highly appreciated! Bind blowing solution

Amazing explanation! Thank you!

Thank you

I am hard time understand why the last solution is linear time. Suppose we have a tree with one left branch of length n-1 and one right branch of length n. Then in order to level the paths you need to run the loop n times, and path length is n. Shouldn't this be O(n^2)?

This is good! Thanks for the post. can you post a variation of this for other 2 LCA problems the other ones doesnt have reference to its parent 1644 1676

Well done mate! Appreciate the thoughtful explanation!

Thanks. bro next time use different colors for the drawing.

No way someone figures this out, but it makes so much sense lol

question: if a node is LCA of itself (descendant) meaning it is the parent of itself so do we have to add the same node as parent of itself instead of Null

amazing work!! :)

At 3:39, shouldn't the time complexity be 2*log(n) instead of 2N? We're only looking at the height of the tree in each case

is it |b-a| = |a-b|?

thanks

Love from India

Tortoise hare strikes again

You need to work on your rationale. Yes you reach the answer but it's not clear why.