In-order Traversal Algorithm | Tree Traversal | Visualization, Code, Example

[@robbie5416]

The video we needed but didn't deserve. Thank you! Never thought a simple video like this would answer so many questions I had.

[@jarjarbinks8954]

this was soooo well done. Thank you a lot. I know Im not the only person who desperately needed this

[@machinelearningandartifici9819]

yes, very well made. really love this concrete dry runs. most cs teachers don't do this in class because of the amount of hard-work it takes to draw this out on a whiteboard... but as a student its so necessary to see this

[@HussainAbdi_]

Thank you! Preparing for my Meta interview and this video is exactly what I needed to help visualise things so I could better understand converting a BST into a DLL. As simple as the code is, being able to visualize recursion as a stack the way you did will really help internalize what's going on for harder problems. Subscribed - I see you're still making videos! Keep creating!!!

[@MyBinaryLife]

ive been trying to learn trees for like a year and this is the best content I've found so far, thank you for putting this out

[@for461]

OMG Sir you are Super. This kind of teaching should be there in every university with visualization. Thanks a lot. Many thanks Sir !!!!! Amazing Explanation.No one could explain like you. You have a lot of Patience Sir !!!!!

[@tosin23]

this is the best explanation video i’ve seen on this topic, thank you!!

[@santiagobenitezperez5240]

this visualization was beyond awesome.. thank you!!

[@MrInboxer]

This video deserves more views and likes!!

[@tanishq2766]

Keep making videos plewaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee, this was sooo good, the visualisation that we needed so damn much! Thankssssssss

[@ImmediatelyLeaveYT]

top notch explanation and content sir

[@cycla]

best visualization of binary trees ever, why aren't you getting more views

[@tbzz99]

What a great video, really really awesome explanation!

[@carefree_ladka]

Thanks a lot . I now understand how recursion works with trees

[@neuralworknet]

Perfect explaination!

[@ojithmagamage1809]

Thanks for this precise explanation!

[@ScraperCross]

Incredibly helpful! Thank you so much!

[@kailashlakshmikanth7386]

great explanation!! :) finally understood tree traversal using recursion

[@asmallapple1509]

Goated video

[@eslamahmed-zr8rl]

This video helped me a lot understanding recursion on trees thanks a lot

[@skylersimpson4844]

bro is the goat

[@shaziakaleem1895]

Great explanation, very helpful. Thank you

[@sekirandahamza1260]

Awesome explanation. You are the best.🙌

[@DP-md4jf]

Wow amazing video thank u sir. Liked and subscribed

[@saliheenafridi9116]

Thank you very much. Yo got a subscriber

[@raghusagar-r3c]

Thank you for the video.

[@venkatraghavansriperambhud1871]

this got me to the closet understanding. Questions how did the tree go from 1 -> 2? Thank you.

[@carefree_ladka]

When the node 1 gets traversed (including its left and right subtree) , it gets popped off the stack and the control goes to as 1 is a left subtree of 2

[@abhitejamandava1619]

Awesome explanation mate

[@TJ_12138]

Very useful!

[@lin99nn]

thanks for sharing, it's very helpful

[@yunanistan2364]

Exactly what I search for.

[@deepamanne]

well done!

[@brettclifford2723]

Omg i love you for this video

[@yashwanth73337]

How does the computer realise that a node has been processed completely i.e left and right child is none...so the current node has to be removed from the stack? How? This is what's been bugging me... How does it realise "it's time remove the node from stack"?

[@ポヤポヤ]

the node from stack will be removed once it reached the 'base case', the base case stops the recursion (you see the code says return). then it will proceed to the next stack.

[@kostiantynshyshkovskyi630]

Perhaps it will be helpful if I try to explain how I understood it. The values are added to the stack(last in first out) and it pops from the stack when we reach the base case. As soon as we face the NULL method returns the value to the caller and then the method will be called with the previous input. For instance, if we have node 1 and there’s no left nor right node it means we have input 1 and try to reach the left child it’s NULL we return the default value and take a step back to the node with 1 value print the value, and we try to reach the right child and it returns the default value because there no right child as well and we again take a step back to the node with the 1 value but we already have the outcome we return it to the caller(the previous one who called the method with this node where the value is 1)

[@gvlog1607]

God bless you.

[@elsayedmohamed451]

many thanks

[@aadityaprakash139]

great visuals

[@kalahari8295]

Omg thank you 😭❤️🔥

[@sash11g]

graet content

[@gokulr3482]

simple but great

[@carolineli7948]

thank you!!

[@arindam6130]

this was helpful, tysm

[@bananaboydan3642]

What prevents the code from going back to 1 when inorder(root.left) is called

[@isaicruz7867]

Think of each call of the inorder functions (inorder(root.left) & inorder(root.right)) causing a “freeze” of the current “instance” that you’re working through. So to your question, when we first visit root 2 (which was executed by calling inorder(root) where root is the value of 2, we skip the base case (of course) and execute inorder(root.left) and we know to replace root.left with the value of 1 since that’s root 2’s left child. At this point, think of the current “instance” that we’re working through (the execution of inorder(root) where root is the value of 2) as now being frozen in time (and for visualization help, think of the white arrow as staying stuck on pointing at the line inorder(root.left)). Since we just executed a recursive call (inorder(root.left)) with root.left being the value of 1, we go back to the top of the function to execute this (inorder(root)) where root is the value of 1. Jumping ahead a few steps since we don’t care about 1 as the root since they have no children, we eventually end up finishing working on inorder(root) where root is the value of 1. So we go back to root 2 where we left off with it “frozen” and the frozen arrow pointing at inorder(root.left). We can now “unfreeze” the arrow since everything we had to do inside of that inorder(root.left) call where root is 1, is now complete. We can now move the arrow down to the next line and continue with the last two lines. Remember, since you specifically asked about root 1, this same “freezing” technique has already occurred to other root values, so you would be “unfreezing” these instances as well when you eventually get done with their left children. Feel free to ask for clarification on anything

[@sameerroshan9542]

what does root mean here int he function definition and base case?

[@juyeon_kim]

thank you

[@srishtijha3]

So helpful

[@encapsule2220]

i still dont get it.

[@semih-ze9uo]

thanks a millions

[@blockchainbaddie]

holy shit

[@DoubleChase]

useful

[@ohsopaid1614]

goat

[@shashicsnitjsr]

How do you create animation ? Do you use powerpoint or some other software?

[@ygongcode]

I used github.com/3b1b/manim and coded all the animations as well as my own data structures for the animations.

[@coffeepwrdcomputers]

@@ygongcode Do you happen to have the source code for the animations available somewhere? I only ask because I just started messing around with manim and example code for making data structure visualizations are few and far between.

[@NavJaswal]

Binary tree questions on leetcode had me cooked fr. NOT ANYMORE.

[@AmanSingh-vy5tk]

The ordin porject (i made it till here)

[@anurag-sachan]

🫡🚀 thanks alot bro.