Same Binary Tree

[@maxbarahona9675]

My man Tushar is the God of programming

[@shankysays]

we can also do it like this. declare a global bool value as true. do any tree transversal and in visit stage make the bool value false as we encounter non equal value. at the end just check the bool value if its still true means the btree are equal or viceversa.

[@jdragon8184]

its almost intership time , wether i get one or not , tushar was a big part of my journey

[@shivtanwar3054]

The way you teach is protean and I liked it so much thanks

[@avinashaitha143]

Welcome back to the Gabage

[@rajesh21602]

return root1.data==root2.data, do we need to change it to if condition, because it would return and end the code at the root it self if they are same and will never get to the recursive part. Please let me know.

[@dineshgarg9567]

Nah, the code is correct. If you read carefully you'll see there's an AND operator between the return statements and the recursive function call statements.

[@prajwalchoudhary4824]

@@dineshgarg9567 thanks

[@madhukiranattivilli2321]

cond1 && cond2 && cond3 would be true only if all 3 conditions are true

[@ManishPandey-mn3gl]

Hey Tushar, Nice solution! you can write it succintly like the following, public boolean isSameTree(TreeNode p, TreeNode q) { if(p == null || q == null) return (p == q); return isSameTree(p.left, q.left) && isSameTree(p.right, q.right) && p.val == q.val; } In the code above, if(p == null || q == null) return (p == q); is equivalent to writing if(p == null && q == null) return true; else if(p == null || q == null) return false; I have a question here, You mentioned that the running time is O(n) here 1) What is n here. Is it total number of nodes in both tree? 2) Shouldn't it be O(min(m, n)) where m = number of nodes in tree1 and n = number of nodes in tree2 ?

[@gokulnaathb2627]

This was so good, Sir

[@bugrahanabdulkarim6034]

or you could write it like this public boolean isSameTree(TreeNode p, TreeNode q) { if (p == null || q == null) return (p == q); return (p.val == q.val) && isSameTree(p.left, q.left) && isSameTree(p.right, q.right); }

[@TheGamerGuy201]

can we write :- if(root1.data == root2.data){ return areIdentical(root1.left,root2.left) && areIdentical(root1.right,root2.right)? }

[@priyankpande9]

Yes it is correct, because in the video if root1.data==root2.data value will be false then it will not go beyond && operation.

[@mystic3662]

Can we also perform in-order and (pre-order or post-order) traversal of both trees and compare if they are equal?

[@bharatgupta3287]

yes but recursion method has less time complexity

[@disenchanted_dove]

There are different tree configurations possible for a single tree traversals results , therefore I don't think it would be right to simply compare the traversals

[@sauravagrawal5013]

@@disenchanted_dove We can compare inorder, preorder and postorder sequentially. If all match, then same.

[@akarshit8556]

@@sauravagrawal5013 It is failed on certain testcases. like [1,1] and [1,null,1].

[@mr_world_wide]

Why can't Two be Two , why does it have to be Thwo

[@pramoth805]

You are little bit faster. Could you please teach a little bit slower

[@rajputtechclasessrtc6876]

challange for u 2,4,6,3,7,7,8,10 create binary tree 😎

[@sauravagrawal5013]

I think U mean BST ?? Not BT. hehehe