Subtree of Another Tree (LeetCode 572) | Full solution with examples | Study Algorithms

[@unemployedcse3514]

guys who ever is confused with '^' in Stringbuilder it is to handle below kind of tree node TreeNode root = new TreeNode(12); TreeNode subRoot = new TreeNode(2); where 12 (tweleve) will be treated as 1,2 subtree as 2 it gives true to figure out item or data in node we are using ^ as seperator still if u need confirmation pls refer test case created by nikhil in description code and nikhil small suggestion pls explain or stick to same logic during dry run , this will give more clarification and reduces doubts , in last example no null were included in ur explanation rest vedio is very helpful thank you ❤

[@alisheheryar1770]

you have to add string("^") to your initial string otherwise it breaks at special cases like s = 12nullnull t = 2nullnull Cheeky Nikhil.

[@garvitarora7777]

thanks bro

[@rahulsinghdeo7519]

Hey can you please explain how the "sb" object of String builder is preserve-ing the value when it moves to next iteration.

[@syedomarali2678]

It won't take O(n). Java's .contains() works in O(MN). Use efficient pattern-matching algorithms like KMP to achieve linear TC.

[@AlexTopGT]

thank you! best explanation.

[@minnikeswarrao6185]

Time complexity shouldn't be O(n) ...since String.contains() takes O(m+n)

[@nikoo2805]

O(m + n) will translate to O(n), where n > m

[@pranavm9301]

O(m+n) is still O(n) since m O(2n) which is worst case when n=m, which is still O(n)

[@smridhjain4100]

Whenever you put StringBuilder sb = new StringBuilder("^"); it passes the test cases but if you dont put the ^ it fails a single test case of root =[1,2] and subroot=[2] , can you explain why please? I think this is the same query that others trying to ask in the comment section

[@nikoo28]

you have to add string("^") to your initial string otherwise it breaks at special cases like s = 12nullnull t = 2nullnull

[@shrirambalaji2915]

Thank you brother

[@killeraloo3247]

OP solution. 🧡 from remote.

[@djpsn7094]

Why we took ^ in StringBuilder

[@nikoo28]

it is just an anchor

[@suryaram6874]

what about inorder traversal?

[@yemeniabedontaclf]

Repeated DNA Sequences. make vedio about this

[@Rieshu-l9i]

Solve "node at distance k from target node" Please🙏

[@pratyushtripathy1463]

Only pre-Order traversal will work here, In other 2 (POT & IOT) all TCs will not pass

[@codeLine7]

very nice explanation.

[@nikoo28]

Keep watching

[@varundwivedi1079]

Great.Thanks

[@blacklight8932]

didnt work for all testcases

[@nikoo28]

Check the code in description. It passes on leetcode

[@adirang]

​@nikoo28 The solution does not work for in order traversal when root is [1,2,3] and sub root is [1,2]. Code returns True but should be False.

[@garvitarora7777]

this is failing in this test case sir Input root =[12] subRoot =[2]

[@nikoo28]

What output are you expecting?

[@sheikhmkrifat7749]

i applied the same way in inorder and postorder after that it doesn't pass all the test cases in leetcode.

[@aryavardhansingh2162]

same here do you know the reason ?

[@sheikhmkrifat7749]

@@aryavardhansingh2162 yeah, i found that preOrder is used for comparing structure of BT, cause here root node comes first , then left and right on the other hand inorder and post dosent follow this, for that preOrder is used to compare two BT.

[@shahriarbadhon4596]

Nice

[@sumeersaifi6354]

you code is not fully correct if there is root 12 and subroot 2 then for 12 it will be 12nullnull and for will be 2nullnull and when you check root.contain(subroot) it will give true but it should be false

[@nikoo28]

it passes the constraints on leetcode. can you explain your test case a little better?

[@sumeersaifi6354]

I need to check will let you know

[@AniketWaghmare-qi7wm]

Hey @nikoo28 Please Make Video On Question Number 110. Balanced Binary Tree. Please Make video fast because my interview in Amazon. And your teaching is so good

[@BACSShaileshShettar]

test cases failing

[@nikoo28]

check the code in the github link provided in video description.

[@adirang]

​​​@@nikoo28The solution does not work for in order traversal when root is [1,2,3] and sub root is [1,2]. Code returns True but should be False.

[@nikoo28]

@@adirang just checked again, code is returning False.

[@droid-aman]

love ur explanion v much,, +1 suscription from my side buddy!!

[@thor1626]

will fail only one test case. Anyways this method is not recommended, good explanation tho.