Binary Tree Preorder Traversal (Iterative) - Leetcode 144 - Python

[@NeetCodeIO]

If you're looking for today's daily LC (Can Place Flowers) -> kzbin.info/www/bejne/kHjboqSgoc-ji6s

[@victoriatfarrell]

For me I think I prefer to append right and left to the stack and just loop while the stack is non-empty, starting by placing the root in the stack before the while loop. Makes it a very clean solution.

[@servantofthelord8147]

I was wondering why he didn't do it that way. It was cool to learn another way though.

[@RakimMiddya]

How does this work?

[@akhileshgotmare9812]

@@RakimMiddya here is the code I think @victoriatfarrell is referring to def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]: if not root: return [] result = [] stack = [root] while stack: cur = stack.pop() result.append(cur.val) # Append right first, so that left is processed first if cur.right: stack.append(cur.right) if cur.left: stack.append(cur.left) return result

[@aadill77]

Exactly I did the same.

[@cyclone760]

I have never thought of the call stack in this way before... great video

[@saraahmed408]

Thank you so much for the explanation. But why the time complexity will be O(log n) if the tree was a balanced binary tree?

[@ashok2089]

It's not the Time Complexity, It's the space complexity he mentioned. So TC is O(n) as we must visit each node. SC is O(h), h is the height of the binary tree, and in the worst case SC could be O(n), in the case of a skewed binary tree. for balanced binary tree SC: O(log n).

[@gsivanithin]

Great explanation! nice

[@theboss-zk4wx]

thank you!

[@gsivanithin]

Easy to understand

[@alexandersmirnov4274]

hi Why did you decide to open a new channel? Where should we expect new videos?

[@chandankhuntia7017]

💞

[@AbhishekAbhishek-dk2pi]

Its getting errors 🤦