Binary Search Tree Iterator - Leetcode 173 - Python

[@lakshyasaharan5348]

Your voice so soothing and smooth.

[@numberonep5404]

ikr!!!!!!!

[@calebi23]

Exactly!

[@DavidDLee]

You CAN do it recursively in Python by writing a Generator function, which uses *yield* and *yield from.* The end result is a very clean in-order recursive traversal that's been consumed sequentially, but with O(h) memory ( O(1) for our own state + O(h) where Python keeps track of the generator state).

[@manishk2429]

Me: The perfect intro doesn't exi- NeetCode: Let's write some more NeetCode today Me: (Throws earphones in the air) THERE IT IS !!!

[@carloscarrillo201]

You're a good man, NeetCode. Thanks!

[@kostaad]

I wonder if a yield based solutions with co-routines will be sufficient for an interviewer.

[@yashshukla1637]

Introduction to Problem: We are implementing a Binary Search Tree (BST) Iterator that allows in-order traversal of a BST. The iterator behaves like standard iterators in languages like Java, ensuring values are returned in sorted order. Three methods to implement: constructor, hasNext(), and next(). Basic Solution: In-order traversal stores all values in an array. hasNext() returns true if values remain, otherwise false. next() returns the next value in the array. Time complexity: O(n) for the constructor and O(1) for hasNext and next. Optimized Approach: Challenge: Reduce space complexity from O(n) to O(h) (h = tree height). New solution will ensure average O(1) time for next() and hasNext() but reduce memory usage by using the tree's structure. DFS and Tree Traversal: The problem can be solved using Depth First Search (DFS) (either recursive or iterative). Recursive DFS stores extra data that can be avoided by using an iterative DFS with a stack. Iterative In-order Traversal: Use a stack to hold nodes, allowing traversal without recursion. Only the nodes from the current traversal path are stored in the stack, keeping the memory usage O(h). Details of the Optimized Implementation: next(): Pops the top node from the stack, checks if the node has a right child, and adds all left descendants of the right child to the stack. hasNext(): Simply checks if the stack is non-empty. Implementation Insights: Memory efficiency: Nodes are pushed onto the stack only when needed (i.e., as we traverse the tree). After processing a node, its right child is explored, ensuring nodes are added in sorted order. Conclusion and Code: The stack and iterative approach guarantee O(h) space complexity. Code for next() and hasNext() is efficient, and repeating 3 lines of logic in the constructor is acceptable for simplicity. Final code successfully passes tests.

[@techandmore12]

Here did it little different: class BSTIterator: def __init__(self, root: Optional[TreeNode]): self.root = root self.stack = [] self.cur = root def next(self) -> int: while self.stack or self.cur: while self.cur: self.stack.append(self.cur) self.cur = self.cur.left self.cur = self.stack.pop() result = self.cur.val self.cur = self.cur.right return result def hasNext(self) -> bool: return self.cur or self.stack

[@7oeseven793]

Is doing the inorder traversal and adding the root to a queue works too? it passes all cases on Leetcode and beats 70%. However, I just seen all the different answers and they all use a stack...nvm just seen more of the video. its too much space complexity.

[@shantanukumar4081]

Great Explanation !!!

[@lucashowes4089]

Ik google prolly pays you a bag but they might be wasting your talents. Haven't had online instruction this quality since khan academy

[@ngneerin]

This comment ruined his life. Now he does his thing for peanuts

[@sprajosh]

@@ngneerin😂😂 I think he’ll figure out what to do with his life

[@eva__4380]

😂😂😂😂😂​@@ngneerin

[@bhanuvarma1239]

In the given question they asked that implement in the ""In-order Traversal""...But you did it in the ""pre-Order"" ##In-order traversal 1.left node 2.root node 3.right node 👉👉Please clear my doubt sir🙏🙏

[@ChanChan-pg4wu]

You are always very helpful! I wish I could work with you in the future!

[@techno_season_14]

Omg! Thank you so much! I love your videos Neetcode! Just curious, could you also solve Perfect Rectangles?

[@xjlonelystar]

Man U doing more leetcode than me and u already have a job ☠️

[@essamobeid1507]

amazing solution! thanks a lot for posting:)

[@alexandremarangonicosta]

Why space is O(h) if the first thing you do is to store the root, which contains all elements in its memory? Looks like space is also O(n). EDIT: you are actually right....the left/right of a TreeNode, in python, only hold the reference of an object, not the object directly...so it behaves like a real pointer in C.

[@usmanakram5191]

Does there exists a solution with O(1) space complexity for this problem?

[@charleskorey6515]

Yes there exists but then you have to do o(n) search everytime and it won’t be an iterator anymore. If you count recursion stack then o(1) is not possible

[@ibrahimalshubaily9520]

Still dont get why does has next work?

[@dewangpatil4990]

Hello Sir i am a big fan......I have a very important request....... Could you please make solution playlist of Striver's SDE Sheet. Its will be very beneficial for us students

[@CommandantNOVA]

A lot of problems on the sheet overlap with the leetcode top 75. There's a top 75 playlist already.

[@dusvn1484]

Ahaahahahah your first appoarch of solution is my solution and I know it's not efficient but I'm still stuct to find better solution.

[@MysteriousIndiaYT]

Bro i am expert in python. But very confused, which career i have to choose for my life. Django Developer vs Data Scientist? Can anyone please guide me. I have little knowledge of both

[@Ankit-hs9nb]

Be a machine learning engineer I am a Data Scientist with 4 years of experience and now switching to Machine Learning engineer

[@sidazhong2019]

@@Ankit-hs9nb The Machine Learning engineer minium requirement is PHD.

[@zohahs5276]

how good one has to be in maths to solve these algos?

[@srx2106]

for some reason the constructor keeps going over my head

[@harshavardhanveerannanaval8605]

Can someone please verify if it could be done this way? class BSTIterator: def __init__(self, root: Optional[TreeNode]): arr=[] def flatten(root): if(root): flatten(root.left) arr.append(root.val) flatten(root.right) flatten(root) self.arr=arr def next(self) -> int: return(self.arr.pop(0)) def hasNext(self) -> bool: return(self.arr) I just done a inorder traversal before hand and store it in an array

[@sidazhong2019]

This problem is easy.

[@adriandelfin8047]

How do I cope with so many failures in a few months

[@fawadazhermalik9885]

Bro failures are an essential part of life that give us new lessons, which we don't get if we don't fail.