Populating Next Right Pointers in Each Node II | Live Coding with Explanation | Leetcode - 117

Detailed explanation for Populating Next Right Pointers in Each Node Leetcode problem 117.
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0:00 Basic BFS
2:40 Follow up [ constant space O(1) ]

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[subhedarsanchay]

This was indeed a tricky problem. The trick of the problem is that we are traversing two levels at a time: Current level and the children of the current level. Introduction of dummy variable in the code before explaining it made it harder for me to understand it. After spending enough time on it, I see it though. If possible, try and match the code with the explanation. I do see that you go back and explain the dummy variable but by that time, viewer is confused already by seeing it in the code. Also, calling the variable with a meaningful name like `childHead` might have made it easier to understand. Good explanation, regardless.

[supreethpadavala]

Wonderful explanation. Felt the explanation is easier than most of LeetCode discussions and official solution. Not to mention it is highly efficient as well. Great stuff guys. Really great.

[xinniu]

This answer is brilliant !!! Thank you for sharing!!! Let me do some explanation as things are not as straightforward for new learners like me:

outer while loop 1: only root is store in queue. The size will be 1, so in the inner while loop we just pop root from the queue and store it in the dummy (dummy = dummy.next). Here node 2 and 3 is added to q.
outer while loop 2: Now we go the next round on the q(2, 3), here dummy is redefined. It first point to node 2, and then in the next inner loop dummy points to 3 and 3 (dummy.next) is pointed to 2 (dummy). Here 4, 5, 7 is added to queue.
outer while loop 3: dummy.next->4, dummy->4; dummy.next->5, dummy(4)->dummy.next(5); dummy.next->7, dummy(5)->dummy.next(7)

[obulpathi]

Excellent solution...better than the leet code's official solution, it was a little bit complex to understand and implement during the interview. thanks for sharing.

[ThePradeep]

I am bit confused on space complexity. As you are creating a new dummy node (from heap space right?) for each outer while loop invocation, you will be creating O(h) dummy nodes (h is height of tree) and hence it is not constant space. Anything I am missing here?

[starpomero]

Could you please tell me why
head = dummy.next;
can get value even though you use p next = node.left ? (it seems that you don’t set value for dummy inside roop)
if we write TreeLinkNode p = dummy; then dummy could reference the p ? so it means both could take same values?

[shreyash]

This is brilliant! Superb explaination.

[mortalphoenix]

I don't understand how dummy node is changing levels without updating dummy node anywhere?

[Rahul-sqgk]

thanks alott.. the leetcode solution hurt my brain. thanks again

[sommayghosh]

Quick and lucid explanation, thank you!

[paragroy]

Nice explanation sir the follow up part is a bit tricky??

[toshanverma]

Very well explained sir. You made the problem really easy. Thankyou sir.

[ZeroTrunks]

Since a new node is being made at every level isn't the space complexity O(h) where h is the height of the tree?

[hasrat]

Thanks I was not able to come up with this solution

[ratnadeepbhattacharya]

You are using the outer while() to traverse the height and inner while() to traverse the level and creating a new node inside the outer while loop. SO in that case you are dynamically creating h nodes where h is the height of the space complexity in that case be O(height) instead of O(1)?

[vincechen]

Thank you so much sir!! You just made my day :)

[FitnessChaos]

Cool video dude. How many years have you been coding by the way to get to this skill level?

[obito]

how its working internally...the code u showed up was easy to understand bt how storing process done of each node?

[Gia]

Space complexity of BFS is O(N), while you should use O(1) in this problem