leetcode 536 Construct Binary Tree from String

[@amijugjug]

Make playback speed to 1.25 and use earphones.

[@ruhigarga9542]

Your way of explaining the logic on the whiteboard and then again via code is super awesome!

[@AdityaSingh-ql9ke]

Awesome, your way of explaining is really what learners newd.

[@mandeep2192]

woww explained so beautifully.

[@mrprime557]

Amazing !!!!!! Love it

[@shaikmohammedrafi3309]

thanks bro for explaning

[@codebix1096]

Happy to help
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[@REGULUS2106]

perfact

[@ritikagupta8847]

What is the time and space complexity?
Time is O(n) and space complexity is also O(n): stack space (in case of skewed tree). Am I correct?

[@vaishaligoyal7478]

Nice explaination

[@codebix1096]

Thanks

[@samratganguly]

Nice Solution Dude . Easy to understand . It helped .

[@samratganguly]

@@codebix1096 and it's time complexity will be O(n) ??

[@codebix1096]

@@samratganguly yes

[@MrThepratik]

precise and well explained

[@codebix1096]

You can subscribe to this channel for more such content 😄

[@niteshverma8281]

@@codebix1096 ruclips.net/channel/UCZJRtZh8O6FKWH49YLapAbQ p

[@ayushkushwaha171]

good explanation

[@codebix1096]

Keep watching
Thank you.
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[@adarshmishra5340]

thanks bro i was missing one case returning root when it is closing bracket though i was getting the correct ans ,as inorder traversal was same but the structure got different..

[@amannegi8256]

are sir sound system thik kro please adhe subs aisi bdjynge

[@snehabaser3155]

How u reach to root.right block.. as upper if condition true and it ovveride root.left.
Please clear my doubt

[@muskanjain4876]

Same doubt

[@lucky_raiser]

Think of recursion friend, you will get out from middle if() loop so the third if() will run , not the first one. So no overriding here.

[@niteshverma8281]

Think of recursion friend, you will get out from middle if() loop so the third if() will run , not the first one. So no overriding here.

[@lucky_raiser]

I think fourth if loop is not required, second if loop will take care of it, correct me if i am wrong.

[@sarveshchavan4391]

No it's required, in fact the if statement on line no 31 is not required, all test cases passed !

[@shrshawn]

this code misses a critical case which is "()" which represents empty subtree for a given root

[@rajatmaheshwari4656]

starts from 1:00

Thanks but what if we want to create a generic/n-ary tree instead of binary tree? 🤦‍♀️ really hard to create..

[@prabhatmishra5667]

Not really, you just have to call recursive function n times in that case with slight modification

[@SandeepKumar-qw3tv]

sound accha karo yaar kuch samujh me hi nhi aa rha hai

[@codebix1096]

Noted.
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[@shashankrustagi2812]

If an input is given to me int string like 123NNNN where N is null, how will you construct the binary tree?

[@AmandeepKaur-ct5mm]

voice too low

[@古小英-u3y]

渐

[@godjames3063]

gg

[@anilchaudhry804]

@@godjames3063 stfu

[@anilchaudhry804]

%&*@###

[@sumitjindal1115]

sir your voice is very low

[@codebix1096]

noted
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