Evaluate Boolean Binary Tree - Leetcode 2331 - Python

[@potansiyel7021]

My friend, I want to examine your brain. After so many leetcode questions, I think you have an extremely superior brain.

[@preritpradeep]

Hey Neetcode, Whenever you get some free time can you please solve the problem - Number of Wonderful Substrings. I haven't been able to find a single good video explaining the intuition of how to solve such problems. Thank you in advance! 😃

[@sreyanth99]

Shoutout to laddoo gang.
Video was helpful as ever, thank you neetcode for the neat code.

[@gabrielvrldev]

Thanks for keep posting the solutions of the daily challenges! It's helping me a lot this past few days!

[@ahmedtremo]

I think the recursive solution is O(log n) as it's a full binary tree

[@deadlyecho]

I solved it iteratively using 2 stacks for simplicity

[@heetvadiya3629]

Never this early, hello neetcode 😊🎉

[@yassine-sa]

You didn't really need a hashmap even if using it doesn't hurt performance much I suppose, I avoided it doing the following:
Whenever I encounter a leaf node, I pop again, if what I just popped isn't leaf, I add the previous leaf and and then add the non leaf, if both nodes are leaf, I pop the operator again and evaluate both of them and append the result. If this wasn't a tree and was some sort of a graph then yes a hashmap would be useful to not evaluate a node multiple times( number of times it's referenced)

[@yassine-sa]

The iterative approach is medium if not quite hard

[@advaitbajaj4241]

Which laddoo do you like bro?

[@priyanshagarwal8490]

2055. Plates Between Candles... Next Please..

[@pastori2672]

the iterative solution is just as efficient as the recursive one right ?

[@NeetCodeIO]

yeah thats right

[@aashishbathe]

LADDOO REFERENCE AGAIN!

[@NeetCodeIO]

🥹🟠

[@scuderia6272]

Ladooo 😂