Level by Level Printing of Binary Tree

[@sandeepannits86]

It is such a relief brushing up things watching your videos one night before interviews :)

[@xVermyy]

Ayyy I'm in the exact same position lol

[@Masa-lo4uw]

same here

[@rahuldevmishra4558]

At 7:08 - The third way - I think the variable currentCount is not necessary. The currentCount holds the count of elements present in the next level. We can get the same value from queue.size() when the levelCount becomes 0 i.e. once the levelCount becomes 0, the only elements present in the queue would belong to the next level. As you clearly explained, as long as levelCount is not 0, remove the element from the queue and add it's children. Once the levelCount becomes 0, then reset the levelCount to queue.size(). I was able to code this without using the currentCount. And thanks a lot for the video.

[@amishashahpatel]

This is really very helpful. This will in fact help solving many problems like Populating Next Right Pointers in Each Node. Thank you so much!

[@daspradeep]

in the first 2 q approach, code can be made simpler by swapping the queues such that you are always operating on the same q

[@tusharroy2525]

Yes that's a good idea too.

[@dibyajyoti_ghosal]

great explanation, it's sad that you don't make videos anymore, maybe apple is not letting you. But your dp contents can be re-uploaded with more explanation on how to approach problems.

[@ANANDJULU]

Before addinga null, we need to make sure 1. we get null returned from queue and also 2. queue is not empty Else it will go in an infinite loop.

[@CEngineer2010]

Around Time 13:27, you forgot Q1.pop and Q2.pop within your inner two while loops. Thank you for posting. Without popping the values from the queue within the two loops. It will create an infinite loop.

[@ankishbansal420]

Thanks Tushar, you are very helpful as always.

[@vgrunert]

Excellent tutorials. Keep up the good work. They really help.

[@DarshanWashimkar]

Why do you need two queue and so many complications? BFS will print desired output. For example, use one queue, visit node(print node), put left and right child in the queue,...pop node from queue.....repeat the process until your queue is empty.

[@debabhishek]

+Tushar Roy even if you want line by line the regular BFS will work.. and I Q is enough.. let me write sudo code. ( and you dont need a delimeter too). let me me write sudo code put root in Q.[ Q has 1 elment root] while ( Q is not empty) { // outer loop int loop_run=Q.size(); for ( i=0;i<loop_run;i++){ // inner loop current node = Q.enqueue(); if ( current_node.left !=null) Q.addelement( curent_node.left); if ( current_node.right !=null) Q.addelement( curent_node.right); print current_node; } // inner loop ends System.out.println(" " ); // look when inner loop ends you have printed 1 level .. // and other level elelements are in Q.so next for loop will print next level. } // outer loop ends * this is very similar to your third approach. ( I have not seen it before commenting it)

[@shubhamvadhera]

+Tushar Roy It can be done using one queue only. Take the size of queue at every iteration and dequeue until current size.

[@abdallaelmedani8933]

Great stuff. Thank you so much for making this video.

[@sumitchohan9473]

It will make more sense if you use word "enqueue" and "dequeue" instead of "push" and "pop" when you work with queues. With this no need to specify if you intend to add/remove them from front/rear.

[@shivanimall3820]

Thank you for the videos. They really help. Can you also make one to find the depth of a node in a binary tree?

[@shivanimall3820]

+Shivani Mall Or would this approach work? while finding the height of a tree, instead of reaching the leaf we reach the desired node and return from there.

[@shivanimall3820]

***** Thank you. That helps!!

[@shivanimall3820]

Also, do you have a video on finding a circular loop in a singly linked list? I am having troubles understanding a part of it.

[@skariaroy5988]

Why can’t you create an array of the same size as the queue and map each tree element to the array like we do in heap. If there is no element we can put null. Then just print from the array by skipping all Nulls?

[@basedmangoes3379]

Awesome video as always. Keep up the good work man.

[@yamanshiekhdeia4778]

Dudeeee! you're a national treasure!

[@kaivulya]

Greaaaatttt!

[@alfabinomial6183]

sir i didnt understand that you said the time complexity is O(n) but this function is working with inner loop ..could you please explain !! i think its O(n^2)

[@HemaLathadaksnamurthymyfamily]

Is this same as creating a linked lists of all nodes at each depth for a binary tree??

[@annastakanova1115]

thanks so much!

[@shobhitranjan3957]

coding in python there is no poll() fun what should i do?

[@נדבסלמן-ט2מ]

Amazing great algorithms

[@SatyendraJaiswalsattu]

great...

[@CraftandDogs]

this is smart approach! Thanks \0/

[@neelamchahal9299]

nice t-shirt

[@magicodabola10]

Thanks for the vid but you are using stacks and not queues. When you say pop and push it's functions of stacks, in queues it's called dequeue and enqueue. Also the "pop" that you are referring to is the LIFO type, which is stacks of course..

[@rashmimishra3031]

Hi Tushar, could you please explain backtracking too.

[@rashmimishra3031]

***** Hi Tushar sorry for confusion. I asked you to explain backtracking in general.

[@SahilSharma-ir7ix]

[@ThexAzox]

ya me chantaron el rojo men :,v

[@VuongGiia]

how about C++??

[@vishalmishra1937]

is it similar to bfs

[@xjsnjkil2070]

I HIT THE LIKE BUTTON BECAUSE YOU ARE A SEAHAWKS FAN.

[@AdityaFingerstyle]

Using a recursive function void display(node* Ptr) { if(Root == NULL) { cout left); cout data right != NULL) display(Ptr->right); } }

[@EternalGlory6]

arkadaşlar lütfen aramızda para toplayıp hocama bilgisayar alalım adam kodu tahtaya yazıyor :'(

[@MohitK96]

could take better example