Lowest common ancestor of two nodes in Binary Tree Algorithm

Excellent explanation 👍

This man is awesome. Thank you.

Can you please tell how is this inorder? Shouldn't it be preorder

What if the root node is one of the two node ??

Explain it for (j,c).... 🤨

I wonder when you first encounter this question, how do you figure out the possible steps for getting the answer? I am having trouble of algorithm questions recently. Very hard for me to think of the answer

I love this guy, very clear explanation !!!

Great Video...
PS-: can watch it at 4x comfortably

Bahut badhiya sir .... Maja a Gaya sir

code doesn't work if one of the node is not present, ideally it should return null but it returns the other node which is found

Thank you

Love you bro. You are so real!!

Clear explanation. Easy to understand

the best!☺

thank you

make more videos plzzzzzzzzzzz :(

Thanks bro

well explained..

Sir, I have watched many other people videos but your's are simply crystal clear

The author is a genius. Thank you very much!

Sir amazing explanation

It would be far better if you also explain the intuition to this solution. Knowing the algorithm makes half of the work but knowing what led to this algorithm will make the solution video totally worthy.

this is how the DSA faculty looks like!

Best explanantion :) .Subscribed

It's good

Should be pre-order instead of in-order.

Great Video. Thank you sir. What happens in the case when there is only one value present in the tree, and the other value is not. In that case, the algorithm is not working.

No need for "else". Great code walkthrough and explanation !

This solution not work when one of the two nodes is found and not the second one. So its return the founded node.

Absolutely love the clear explanation and walkthrough. Thank you!

OKK .. lets just appreciate his innocent smile!

nice video.............................

great explanation but i think this is pre order traversal

Thanku ji

this solution will not work in case if B==root->node ans c is not present

sir this is a very good explanation, just one point that you should have told that we are taking the assumption that both the nodes exist in the tree

Thank you for your tutorial. You are one of the best so far. For the if(left !=null && right = null) return root. Is that right or (right !=null) .Please explain. Thank you very much.

thanks for making things clear and simpler sir

Sir ji, apka teaching ka FAN ho gya hu me to .. love you.
Some of the best things about your video,
1. your slow explanation.
2. code walk through
3. Complete explanation with lots of examples.

Ur a amazing teacher

usually i have a hard and long time understanding these kinds of videos, but your videos are so clear I feel like even a toddler can understand

so clear explanation!

really good method of explaining!!

This won't work when one of the two nodes is not found.

Hello, Thanks for explaining so well, In your logic for finding out LCA works fine if both the nodes are present and both the nodes are part of different sub tree. Two situation does not handle in this code. 1) if either of the node is not present in tree itself 2) if second node is part of subtree of first node. for exm, P and L nodes. If user enters these two nodes then this algorithm will not work.
Solution algorithm:
1) Both the nodes must present in tree otherwise print error and return.
2) if left and right returns non NULL, this is LCA. Print LCA and return NULL.
3) if second node is in subtree of first node then only first node we will return till root node. in such case main function shall have a piece of code, if final return is non NULL then this is the LCA, print LCA.
I can post the code if someone needs. thanks

DAMM !!! ABSOLUTLY SPOT ON

You are the best teacher sir thank you so much for all this solutions

try this approach
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {

if (p->val<root->val && q->val<root->val)
return lowestCommonAncestor(root->left,p,q);
if (p->val>root->val && q->val>root->val)
return lowestCommonAncestor(root->right,p,q);
return root;
}

i was not able to understand this question's solution from any other video but your simple explanation worked and i will never forget this solution again, thanks so much sir!