Check whether BST contains Dead End | BST | Love Babbar DSA Sheet | Amazon🔥

[@rrr9848]

Maja a gaya .. sir 🎉

[@DurgaShiva7574]

You are a coding prodigy 🎉

[@a3rdtierguy864]

MAINE BHI PAHLA APPROACH USE KARKE SOLVE KARNE KI KOSHISH KIYA BUT TESTCASE FAIL HO JAYEGA IF BST KEVAL EK SIDE ME JA RAHA HO

[@ridimabajpai4911]

Hey!! I have been following your videos and they are great.... Superb explanation... Just a request plz upload largest bst video soon

[@sagartaak4017]

2-1 = 3 and 2+1 =4 noice mazak se hat kar thanx for your content bro love and respect from Himachal !!!!

[@altafmazhar7762]

Nice Bro good to see you are going from brute force to optimal

[@AbhishekMishra-me4mu]

Can we do this by calculating inorder successor and inorder predecessor of all the leaf nodes and comparing each time if (inorder successor of leaf node i) - 1==leaf node i==(predecessor of node +1) then return true; ??

[@Priyasingh-xr6df]

why a new node is always is added at the leaf node? we can also insert a new node at internal nodes also which has only one child. In this question the first example they have given, we can insert 3 after 2, which is an internal node.

[@carry1449]

you may define a dead end as a Node that will never ever have even a single child

[@srinivaskonduri7496]

Great video.

[@shiveshpandeyiiitbhopal5463]

Is 7 not the dead-end in the example you are showing?? Because we can't insert any node to it's right.

[@ranvijaypatel3213]

7 is not a end (leaf) node ...

[@Ajit-p9h]

why a node having single child can't be a dead end? In your exmple 7 can also be a dead end.

[@TekaneChaitanya]

Dusare channel par dsa ke videos nahi aa rahe jaldi c++ ke

[@CodeLibrary]

It will come in all alternate days

[@rdxgaurav3483]

2-1 = 3 Haha , kabhi kabhi expert minds bhi hil jaate hai

[@The_Gurugram_Voyager]

Why this video got deleted yesterday?

[@CodeLibrary]

Because by mistake I said the wrong time complexity in the second approach

[@memefulland7140]

Whats wrong in this approach // int f; // void solve(Node* root,Node* p,Node* q) // { // if(!root) return; // solve(root->left,p,q); // if(p && p->data-1==q->data && p->data+1==root->data && !p->left && !p->right) // { // f=0; // } // p=root; q=p; // solve(root->right,p,q); // } // bool isDeadEnd(Node *root) // { // f=1; // Node* p=NULL; // Node* q=new Node(0); // solve(root,p,q); // return f; // }

[@kritikadwivedi7736]

Thanks code with harry

[@mdhaidarparwez968]

7 is also a dead end

[@rrr9848]

No

[@rrr9848]

No

[@prakharsahu7145]

get to the point get to the point !!!!!!