Binary Tree - 45: Print Root To Leaf Every Path In Binary Tree

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Source Code:thecodingsimpl...
Solution:
- We'll keep on adding the node value in array & whenever there is leaf node, we'll print the array starting from 0 to current index
- Time Complexity: O(n)
- Space Complexity: O(n)
This Problem is synonym of following problems:
binary tree print every path from root to leaf,
print root to leaf path in binary tree,
coding simplified
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Опубликовано:

30 сен 2024

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Комментарии : 18

@sivaganesh4489 4 года назад

Most underrated channel

@CodingSimplified 4 года назад

Thanks. Please share the channel content with your friends.

@sivaganesh4489 4 года назад

@@CodingSimplified defently already suggested your binary tree playlist to many of my friends please do more videos

@CodingSimplified 4 года назад

@@sivaganesh4489 Thanks.

@upendrachauhan5260 Месяц назад

do not waste your time by watching this video

@jabedakhtar5156 4 года назад

wish I could know this channel before ❤️ lovely explanation brother ❤️

@CodingSimplified 4 года назад

Thanks for your nice feedback. Keep Watching.

@sojanmathew5875 3 года назад

How about using stack instead of array? (no need to manage index). If ( node is leaf) { print stack; pop node; return}

@ShreyaSingh-vr9qi 4 года назад

How we can solve this problem using iteration ??

@CodingSimplified 4 года назад

Here we're solving via preorder. You can take a cue from following video, where we're executing preorder without recursion. : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-qv1nLlk5LEU.html

@tastaslim 4 года назад

Easily using Iterative preorder and a map to have a check of nodes visited or not.

@soumavanag5025 3 года назад

thank you very much :)

@CodingSimplified 3 года назад

Thanks for your nice feedback. Keep Watching.

@selvalooks 4 года назад

cool , i had the solution of removing the leaf node from the array after printing, but your solution is better its not updating the index++ instead index+1, nice !!!

@CodingSimplified 4 года назад

Thanks for nice feedback. Keep Watching.