At around 4:39 I think "if not self.value:" will only work as intended if we already know that we deal only with non-zero numbers(not 0 and 0.0) and non-empty iterables(not ( ,), [ ] and "") as they all will evaluate to False and we will be replacing that 0(or whatever) with value but not None which is the default and we don't want. In most cases, the "if not x:" and "if x is None:" structures are interchangable(again, non-zero numbers and non-empty iterables) but you really need the understanding when to use which one Edit: for the iterables, it could work if we make some custom parent class of the iterable which we implement comparing methods in, but what I mentioned before still applies to numbers
I don't understand how your traversals ever printed anything but the most left node? It's clearly my thinking thats wrong, I'm aware, but you never go back up one after printing, so how does it happen?
Hey, I think I can help you if you still don't understand it. Think of recursive functions like this: each time that the function gets called but hasn't made it to the last line of code, there is a function that hasn't been completed yet. That function will be waiting in the computer's memory for its time to continue where it left off. In his example where the root node is 10, by the time we get to Node 1 there are 4 unfinished function calls. Node 1 is the first node where self.left is no longer True or in other words where a left node doesn't exist. We print the value of the current node which is 1 and continue to check if there is a right node. There is no right node so that essentially terminates that function call. We then return to the most recent function call from Node 2. Upon returning to this function call we continue where it stopped. We had already checked if the left node exists so the next line of code gets executed which is the print statement. This prints the current value which is 2 and now we check if a right node exists. A right node does exist so we call the inorder traversal function on the right node. Left doesn't exist for node 3 so we print 3 to the terminal. We check if the right exists and it doesn't so that terminates that function call and we return to the next incomplete function call which is node 4... Hope this explanation helps
No BS, no time wasting, no unnecessary talk. Straight to the point. Beautiful. W channel.
what bs mean?
There is literally no one who has taught this concept in python this easily so far in my knowledge.......thnank you
YES! Literally!
I love how you do things fast and don't make a big deal about writing good code
I never think that anyone teach BST so fast and clearly, but he is.
At around 4:39 I think "if not self.value:" will only work as intended if we already know that we deal only with non-zero numbers(not 0 and 0.0) and non-empty iterables(not ( ,), [ ] and "") as they all will evaluate to False and we will be replacing that 0(or whatever) with value but not None which is the default and we don't want. In most cases, the "if not x:" and "if x is None:" structures are interchangable(again, non-zero numbers and non-empty iterables) but you really need the understanding when to use which one
Edit: for the iterables, it could work if we make some custom parent class of the iterable which we implement comparing methods in, but what I mentioned before still applies to numbers
Literally best video ,i have watched like 10 other channels. i understood the best with this one! Thank you!
better than anything out there on the internet. great work man
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If you had posted this video 5 days ago. I would for sure be able to pass my exam on algorithm :c
14:25 Why do you suddenly need to return the function call? Isn't it enough to simply call the function like you've done up until this point.
Thank you so much, this was really helpful! Spent ages trying to work this out before coming across this video)
Maaaan ! God Bless You . May Your Loved Ones Live Forever Alongside with you
Awesome video finally a good one that helped me grasp the concepts. Thank You😊
W video i watched a ton of videos on search trees and this is the only one that made complete sense!!
Thank you! For greener like me it was so much understandable!
thank you so much for this man, respect from Algeria
Finally a tutorial on how to actually use this thing.
thank you for the video! it was really helpful to get the grasp on it
Thanks for watching!
Thank you, your explanation was super clear and super helpful!
Finally found amazing explanation of binary tree implementation. Wow
Thank you so much! Absolute godsend!
you should be teaching CS at MIT dude! amazing video!
You are the BESTTTTTTTT‼
Amazing Vid!
Nice, thx for all your hard work making tutoring videos
Your Video was Very Helpfull...
Great help cheers mate
Thank you for providing value
Great...good job!
This guy is the goat
you are a very awesome teacher
well explained
Thank you sir.
Awesome bro
I don't understand how your traversals ever printed anything but the most left node? It's clearly my thinking thats wrong, I'm aware, but you never go back up one after printing, so how does it happen?
Hey, I think I can help you if you still don't understand it. Think of recursive functions like this: each time that the function gets called but hasn't made it to the last line of code, there is a function that hasn't been completed yet. That function will be waiting in the computer's memory for its time to continue where it left off. In his example where the root node is 10, by the time we get to Node 1 there are 4 unfinished function calls.
Node 1 is the first node where self.left is no longer True or in other words where a left node doesn't exist. We print the value of the current node which is 1 and continue to check if there is a right node. There is no right node so that essentially terminates that function call.
We then return to the most recent function call from Node 2. Upon returning to this function call we continue where it stopped. We had already checked if the left node exists so the next line of code gets executed which is the print statement. This prints the current value which is 2 and now we check if a right node exists. A right node does exist so we call the inorder traversal function on the right node.
Left doesn't exist for node 3 so we print 3 to the terminal. We check if the right exists and it doesn't so that terminates that function call and we return to the next incomplete function call which is node 4...
Hope this explanation helps
Great man .......
How will we delete a node from it
Can binary search tree have duplicate values??
Yes, I think it would go to the right of its parent node
Thanks
makasi bang
He codes so fast 😮
Perfect
If I run your code is error
yeh same
same
The non binary network diagram on the thumbnail was triggering me 😂
You are a G
Nacho Varga
Lol kind of same but he is certainly smarter XD
waiting for the linklist
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