Balance a Binary Search Tree | Simple Explanation | Leetcode 1382 | codestorywithMIK

[@ankitrajpootr]

Today is my Birthday🎉.

[@venkatarohitpotnuru38]

happy birthday bro...

[@codestorywithMIK]

Wish you a very very Happy Birthday Ankit ❤️❤️❤️ Guys let’s all wish him and make his day 😇🎉🎊🎁🎈🎂

[@AbhijeetMuneshwar]

Happy Birthday, Ankit 🍰🎉

[@varunpalsingh3822]

Happy Birthday bro, Enjoy your day 🎊🎉

[@Himanshh_29]

Happy Birthday Ankit 🥳

[@AbhijeetMuneshwar]

Respected MIK Sir, Leetcode publishes new problem of the day at 5:30 AM daily. You've published this video just after an hour. Hats off to your dedication and efforts. 🙏🙇 You're an inspiration ⚡

[@codestorywithMIK]

Means a lot ❤️🙏

[@vishalkurve93]

​@@codestorywithMIKReally sir you are helping us a lot🙌🙌🙌🥹

[@gui-codes]

Banda legend hai yaar. Itna determination chaie bas life me.

[@kaustubhchaudhari9838]

First shortest video of MIK bhai ever seen

[@rajatrajgautam3224]

but bhai worth it h

[@gui-codes]

ha ha sahi baat. but video is gem as always.

[@molyoxide8358]

Go see his shorts, even more shorter. Just joking buddy.

[@adsnehi]

I am not able to solve any daily challenge problem. but after watching 25% of you video I am able to solve it

[@krunalkp1032]

Hi Mik you are really doing a great job you way of teaching is really nice, I code in python hence i watch your all video till the coding part and am able to do it by my self. Thanks much recursion is indeed work magically. Python Code : class Solution: def balanceBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]: arr = [] def inorder(root): nonlocal arr if root is None: return inorder(root.left) arr.append(root.val) inorder(root.right) inorder(root) def makebalance(left,right,arr): if left > right: return None mid = left + (right - left) // 2 root = TreeNode(arr[mid]) root.left = makebalance(left,mid-1,arr) root.right = makebalance(mid+1,right,arr) return root left = 0 right = len(arr) - 1 return makebalance(left,right,arr)

[@rohan8758]

Awesome explanation bhaiyan, You are like a angel to me who is leading me towards to crack FAANG or product based companies interview.

[@Lakshya-f4l]

"Recursion leap of faith" . . . Thankyou sir!

[@abhinaysahuu]

sir aapki explanations really the best h. mene ek striver ki binary tree ki video dekhi, smjh ni aaya code, aapki video dekhi, easily aa gya. please keep up the good work sir !

[@manasdeora4601]

Sir, you have been helping many like me understand the concepts in deep. Thankyou so much for your efforts.

[@CarWarYaar]

Bhaiya aap bhot acha samjhate ho , dsa ke saare patterns pe koi video bana dijiye , question identify krne me problem hoti hai mostly

[@Abhishek-y6i3j]

short and crisp explanation thanks for this .

[@manojitsaha6262]

AVL Tree 🙂 Mujhe college ka kuch samajh nahi ata apke video se samjha ye concept acche se ❤❤❤

[@raunakgiri5033]

It would be very helpful, if you also provide contest video solution, specially only problem D. Thanks.

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤❤

[@ReshuSharma-jc7yu]

Bhaiya your segment tree playlist is awesome and looking forward for your backtracking and heap playlist

[@meesalamahesheee_0056]

Sir, how did you gain so much knowledge and become consistent in DSA? What experiences and difficulties did you face during your initial stages in DSA, please tell?

[@aws_handles]

Guruji 🙌🙏🏻

[@gui-codes]

MIK bhai zinda baad

[@bhuppidhamii]

East ho ya West, MIK is best.

[@venkatarohitpotnuru38]

bhaiyya please do problems on partition dp

[@codestorywithMIK]

Soon planning in dp concepts playlist ❤️

[@nish0798]

@@codestorywithMIK also please try to give tips how to be good at bottomup and when are u planning to do dp concepts and how amny videos estimated

[@venkatarohitpotnuru38]

@@codestorywithMIK Thank you bhaiyya..

[@ropurifiedwater]

@@codestorywithMIK thank you sir

[@FanIQQuiz]

Resursion ❌ Kahani (story) ✅

[@aizad786iqbal]

awesome

[@dayashankarlakhotia4943]

Good explanation 👏 👍

[@AdarshSingh-is6mg]

bhaiya aap sheet recommend krskte ho koi for interview prep.

[@Coder_Buzz07]

"College mein yeh problem karaya hoga" - college mein kuch nhi karata bhaiya sab khudse karte hai hum log😂😂

[@kartikforwork]

karwata hai na clg, maths, phy, chem, dip, human values, tecnical communication ... me time waste

[@Coder_Buzz07]

@@kartikforwork 😂🤣🤣sahi bole bhai

[@imPriyansh77]

Thank you bhaiya

[@jeehub041]

Sir jo new tree hum log construct kar rahe usko bhi toh space complexity me consider karenge na?

[@pokeindia5361]

Bhaiya please wo palindromic substringa ka blue print aapne diya tha uske variant karwao please..... Abhi placements aa rahe h

[@dhairyachauhan6622]

This is how i did it. was wondering, how to do it inplace, i know AVL TREE but never implement the rotation. Even after being decent in recursion, it is difficult for me to implement it 😅 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: vectorinorder; void fillInorder(TreeNode* root){ if(root == NULL){ return; } fillInorder(root->left); inorder.push_back(root->val); fillInorder(root->right); } TreeNode* makeTree(TreeNode*root,int start,int end){ if(start > end){ return NULL; } int mid = start + (end-start)/2; TreeNode * node = new TreeNode(inorder[mid]); node->left = makeTree(root,start,mid-1); node->right = makeTree(root,mid+1,end); return node; } TreeNode* balanceBST(TreeNode* root) { fillInorder(root); return makeTree(root,0,inorder.size()-1); } };

[@rhydhamsingla9597]

Bhaiya please make a separate video on DSW algorithm. I wanted to learn it and there is no video tutorial available. I checked those which were available, they don't explain any near to you. Please help!

[@bhaktisharma9681]

Mujhe BST ki playlist he nahi mill rahi hai, koi pls share kar sakt hai uska link?

[@piyush1394]

class Solution { public: TreeNode* balancedTreeRoot(int l, int h, vector &arr){ if(l > h){ return NULL; } int mid = l + (h-l)/2; TreeNode* root = new TreeNode(arr[mid]); root->left = balancedTreeRoot(l, mid-1, arr); root->right = balancedTreeRoot(mid+1, h, arr); return root; } void createArray(TreeNode* root, vector &vec){ if(root == nullptr) return; createArray(root->left, vec); vec.push_back(root->val); createArray(root->right, vec); } TreeNode* balanceBST(TreeNode* root) { vector arr; createArray(root, arr); return balancedTreeRoot(0,arr.size()-1, arr); } };

[@sabbyasachichakraborty1824]

LC 3193 kara do pls. Baaki jagah se samjh hi nhi aa rha. Count inversion wala

[@rahulnegi4027]

us bro us

[@focusman8801]

@codestorywithMIK Bhaiya please count inversion wala bana do

[@10minutes_cs]

yes please, male video on this.

[@__ankush_kushwaha]

❤❤❤

[@top_g755]

Sir segment tree plz

[@Rajdweep]

man i didn't get the intuition for this

[@yogeshchauhan9401]

In the if u are checking nums[idx]

[@theUnkown617]

arre bhaiya DSW bhi padha dete na 😭 me vohi padhne aya tha pr apne dokha de diya

[@mayank1522]

yaha par l > r kese hoga , l increase kese ho rha h ya r decrease kese ho rha

[@yogeshchauhan9401]

We are recursively passing them in the function like for left subtree construct(l,mid-1) and for right subtree(mid+1,r) and as a base case we are checking is l>r that means there is no more element to process so we return NULL.

[@ManojKrVerma-vw4dx]

What is the issue with this code ? /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: void helper(TreeNode* root, vector &sorted) { if (root == NULL) return; helper(root->left, sorted); sorted.push_back(root); helper(root->right, sorted); } TreeNode* balancedBST(vector sorted, int l, int h) { if (l > h) return NULL; if (l==h){ return sorted[l]; } int m = (h+l) / 2; TreeNode* root = sorted[m]; root->left = balancedBST(sorted, l, m - 1); root->right = balancedBST(sorted, m + 1, h); return root; } TreeNode* balanceBST(TreeNode* root) { // first get sorted vector vector sorted; helper(root, sorted); //return NULL; return balancedBST(sorted, 0, sorted.size()-1); } };

[@nish0798]

public int subarraySum(int[] nums, int k) { // Call the helper function with an initial sum of 0 return helper(nums, k, 0); } private int helper(int[] nums, int k, int idx) { if(idx>=nums.length) { if(k==0) { return 1; } return 0; } int take=0; if(nums[idx]

[@nish0798]

@codestorywithMIK please reply

[@molyoxide8358]

Bro my code is not running pls help me: class Solution { public: int n = 0; vector arr; TreeNode* newRoot = NULL; TreeNode* temp = NULL; void SolveLeft(int l, int r, vector &arr) { if(lleft = new TreeNode(arr[mid]); temp = temp->left; SolveLeft(l, mid-1, arr); } } void SolveRight(int l, int r, vector &arr) { if(lright = new TreeNode(arr[mid]); temp = temp->right; SolveRight(mid+1, n-1, arr); } } void Inorder(TreeNode* root) { if(root != NULL) { Inorder(root->left); arr.push_back(root->val); Inorder(root->right); } } TreeNode* balanceBST(TreeNode* root) { Inorder(root); n = size(arr); int mid = (0+n-1)/2; newRoot = new TreeNode(arr[mid]); temp = newRoot; SolveLeft(0, mid-1, arr); temp = newRoot; SolveRight(mid+1, n-1, arr); return newRoot; } };