Binary Search Tree Interview Problem | 5 Binary Search Tree Problems in One Shot

[@CoderArmy9]

Chamko do Binary Search Tree ko bhi....❤

[@divyanshrawat2859]

💙

[@push_3872]

chamak gaya

[@allinonemoviesyt]

1:24:21 Homework - Time Complexity : O(N) hogi kyoki array ke har element ko ek baar traverse kiya hai Space Complexity : O(H) hogi H tree ki height hai , ye space hamara call stack lega Node* createBST(vector& nums, int s, int e){ // agar array me koi element hi na bacha ho to return krdo NULL if(s>e) return NULL; // array ka middle element find kro int mid = s+(e-s)/2; // jo middle element aaya hai usko root banao Node *root = new Node(nums[mid]); // array ke left part ko root ke left me daaldo root->left = createBST(nums,s,mid-1); // array ke right part ko root ke right me daaldo root->right = createBST(nums,mid+1,e); // phir root ko return krdo -- baaki ka kaam recursion khud krdega return root; }

[@AbhishekKumar-lp7wy]

Nice explanation ASHU ❤

[@allinonemoviesyt]

@@AbhishekKumar-lp7wy thanks abhi

[@OK-ku8ez]

Array to BST creation : TC = O(N) : Each node will be traversed SC = O(H) : log(N) Node* ArrayToBST(vector&nums, int start, int end) { if(start>end) return NULL; //Create Node int mid=(start+end)/2; Node*temp=new Node(nums[mid]); //Left Child temp->left=ArrayToBST(nums, start, mid-1); //Right Child temp->right=ArrayToBST(nums, mid+1, end); //Return address return temp; } void preorder(Node *root, vector&ans) { if(!root) return; ans.push_back(root->data); preorder(root->left, ans); preorder(root->right, ans); } vector sortedArrayToBST(vector& nums) { // Code here vectorans; Node *root=ArrayToBST(nums, 0, nums.size()-1); preorder(root, ans); return ans; }

[@VineetDixit]

same brother

[@ankushladani496]

Kth largest ka logic pehle hi soch liya tha Bhaiya...❤ Thank You...✅🚀

[@beingbillionare2030]

thankyou bhaiya ethna mast content ke liye jo free diye hai aap

[@ParvezKhan-mj7yt]

Like kar do sab log. He needs to be motivated❤

[@fachoyt9686]

Thank you so much for great explanation. My all doubts have been cleared after watching this Lecture.

[@rohitbisht1107]

Bhaiya web development ki bhi series laega apke padhane se sare doubt clear ho jate hai❤❤

[@heetpatel3037]

Thankyou Ek dum badhiya content bhaiya thank you

[@sumitvishwakarma1075]

Happy Mahashivratri Bhaiya ❤ Sab Chamak gaya ❤

[@ankushladani496]

Array to BST pehle hi kar diya Kal video nahi aaya tha to khudse 5 problem solve kiye the... 💯✅

[@vivektechtuber204]

Happy Shivratri Bhaiya Har Har Mahadev 🙏🙏

[@CoderArmy9]

Happy Shivratri Bhai❣

[@RJFF3-r7w]

Array to height balanced BST. T.comp = O(N),bcoz , all nodes will be created. S.comp=O(H) or we can say Log(N) due to recursive call. Node* sortedArrayToBST(vector& nums) { // Code here int n=nums.size(); Node *temp=create(nums,0,n-1); return temp; } Node *create(vector&nums,int start,int end) { if(start>end) { return NULL; } int mid=start+(end-start)/2; Node *temp=new Node(nums[mid]); temp->left=create(nums,start,mid-1); temp->right=create(nums,mid+1,end); return temp;

[@OK-ku8ez]

Check BST : Upvote if you liked it bool BST(Node *root, int &prev) { if(!root) return true; return BST(root->left, prev) && (root->data > prev ? prev=root->data : 0) && BST(root->right, prev); } bool isBST(Node* root) { //Optimal approach(2) - My method int prev=INT_MIN; return BST(root, prev); }

[@chanchalsoni7535]

this course is literally ....GOD LEVEL🥵.....unmatchable bruhh

[@rishusachan2618]

Convert Sorted Array to Binary Search Tree class Solution { private: TreeNode* constructBST(const vector& nums, int left, int right) { if (left > right){ return NULL; } int mid = left + (right - left) / 2; TreeNode* node = new TreeNode(nums[mid]); node->left = constructBST(nums, left, mid - 1); node->right = constructBST(nums, mid + 1, right); return node; } public: TreeNode* sortedArrayToBST(vector& nums) { return constructBST(nums, 0, nums.size() - 1); } };

[@PankajBabu03]

GOOD MORNING BHAIYA ❤❤❤❤❤❤❤❤

[@CoderArmy9]

Good Morning bhai❤

[@nikhil316_pspk]

balanced BST ....... i think time complexity is O(n) and space complexity O(height of tree) bhai ek like kardo na #include using namespace std; #include class Node{ public: int data; Node *left,*right; Node(int val){ data=val; left=NULL; right=NULL; } }; Node* merge(vector&v,int start,int end){ if(start>end) return NULL; int mid=start+(end-start)/2; Node* temp=new Node(v[mid]); temp->left=merge(v,start,mid-1); temp->right=merge(v,mid+1,end); return temp; } void preorder(Node* root){ if(!root) return; cout

[@asad_iliyas]

H/W = Solution T.C. = O(N) creating each node by traversing each one of them. S.C = O(h) or O(N) recursive call will use stack memory and in worst case height of tree will be equal to N. class Node{ public: int data; Node *left,*right; Node(int value) { data = value; left = NULL; right = NULL; } }; class Solution { public: Node* BalancedBST(int start,int end,vector& arr) { if(start>end) return NULL; int mid = start + (end-start)/2; Node *temp = new Node(arr[mid]); temp->left = BalancedBST(start,mid-1,arr); temp->right = BalancedBST(mid+1,end,arr); return temp; } void preorder(Node *root,vector&ans) { if(!root) return; ans.push_back(root->data); preorder(root->left,ans); preorder(root->right,ans); } vector sortedArrayToBST(vector& nums) { // Code here Node *root = BalancedBST(0,nums.size()-1,nums); vectorans; preorder(root,ans); return ans; }

[@nepalihoni9213]

sc = log(n) / o(h), SC can never be o(n) in any worst case

[@rishabhshenoy3258]

2:04 galat h because right side of left ko root wale se bhi compare karna hoga

[@nandanks2068]

40:19 We can initialize prev=INT_MAX;

[@chanchalsoni7535]

homework que ka ans TreeNode* solve(vector nums,int s,int e){ if(s>e)return NULL; int mid=s+(e-s)/2; TreeNode* temp=new TreeNode(nums[mid]); temp->left=solve(nums,s,mid-1); temp->right=solve(nums,mid+1,e); return temp; } TreeNode* sortedArrayToBST(vector& nums) { int n=nums.size(); int s=0; int e=n-1; TreeNode* ans= solve(nums,s,e); return ans; } };

[@DarkConfidence-zl1xm]

Hello bhiya i am doing bca from a random private collage i am in my 1st year but i have completed c and c++ completely and currently following your dsa series i am very glad to found a teacher link you i understand all the concept of dsa very easily thanku so much can you please tell me if i do hardwork so can i get good job after bca without doing mca.

[@kartikay0111]

LeetCode 108 for Homework class Solution { public: void create(TreeNode* &root,vector &nums,int start,int end){ if(start>end) return; int mid= (start+end)/2; // TreeNode* node = new TreeNode(nums[mid]); root = new TreeNode(nums[mid]); create(root->left,nums,start,mid-1); create(root->right,nums,mid+1,end); } TreeNode* sortedArrayToBST(vector& nums) { TreeNode* root; create(root,nums,0,nums.size()-1); return root; } }; Time complexity = O(n) Space complexity = O(H) -->height of tree

[@Harshmishra0007]

Ram ram bahi❤

[@CoderArmy9]

Good Morning bhai❤

[@Sanataniblood-mb5pu]

Funny part is really funny bhaiya 😂😂

[@Aniruddha_Mukherjee]

Good morning bhaiya ji😍

[@CoderArmy9]

Good Morning bhai❤

[@ankushladani496]

Done Bhaiya... Day 165/180 ✅

[@vaishalichoudhary5]

HOME WORK:) CREATE BST USING SORTED ARRAY: class Node{ public: int data; Node *left; Node *right; Node(int a){ data=a; left=NULL; right=NULL; } }; class Solution { public: Node* BST(vector&ans,int s,int e){ if(s>e){ return NULL; } int mid=s+(e-s)/2; Node *temp=new Node(ans[mid]); temp->left=BST(ans,s,mid-1); temp->right=BST(ans,mid+1,e); return temp; } vector sortedArrayToBST(vector& nums) { int s=0; int e=nums.size()-1; Node *head= BST(nums,s,e); } }; TIME COMPLEXITY:O(N) SPACE COMPLEXITY:O(N)

[@Pallab01]

5: Kth Smallest Element: void kSmallest(TreeNode *root,int &small,int &k) { //base case if(!root) return ; //LNR //left side kSmallest(root->left,small,k); if(k>0) { small=root->val; k--; } else return ; // right side kSmallest(root->right,small,k); } int Solution::kthsmallest(TreeNode* A, int B) { int small; kSmallest(A,small,B); return small; }

[@Pallab01]

3 .Sum of k smallest elements in BST void sumBst(Node*root,int &ans,int &k) { if(!root) return ; //left side sumBst(root->left,ans,k); //sum of k smallest element if(k!=0) { ans=ans+root->data; k=k-1; } else return ; //right side sumBst(root->right,ans,k); } int sum(Node* root, int k) { // Your code here int ans=0; sumBst(root,ans,k); return ans; }

[@Chandan_agrawal]

my logic: root node ke left child ke right most node and right child ke left most node ki value se bst check karenge in question no 1. but apka logic jyada simple hai find inorder traversal and go on.

[@ankushladani496]

Thanks Bhaiya...❤🎉

[@AMITKUMAR-ds4hp]

Good morning boss ❤❤

[@CoderArmy9]

Good Morning bhai❤

[@ParvezKhan-mj7yt]

Check BST. Why create inorder fast. Can't we check directly through recursion

[@YashSaini007]

Bhaiya Radhe Radhe 🙏

[@CoderArmy9]

Radhe Radhe bhai

[@relaxingtime2411]

Problem 3 :- void inorder(Node *root, int &sum, int &k) { if(!root) return; inorder(root->left, sum, k); if(k>0){ sum+=root->data; k--; } inorder(root->right, sum, k); }

[@sdtedit5165]

kth largest using inorder and stack void elements(Node* root, stack&ans){ if(!root) return ; elements(root->left , ans); ans.push(root->data); elements(root->right, ans); } int kthLargest(Node *root, int K) { //Your code here stackans; elements(root, ans); int k2=K-1; while(k2--){ ans.pop(); } return ans.top(); }

[@mohit6215]

Bhai aapki health kaise hai please reply🥰🥰

[@Aniruddha_Mukherjee]

Problem 3 : void getSum(Node* root , int &k,int &sum) { if(root==NULL) return; //left most node tak ja rahe hain. kyunki hume chota elements left side me hi milta hain BST mein getSum(root->left, k , sum); /* ye if condition chalega jab left mein jate jate NULL mil jayega. aur ye bhi check kar rahe hain , ki kya hum K smallest elements dekh chuke hain ? agar k==0 ho jata hain to hum tree me ghumna bandh kar denge.*/ if(k>0) { sum+=root->data; k--; } //agar k==0 , matlab K smallest elements mil gaye toh return kar rahe hain. else { return; } // right subtree ki aur ja rahe hain. getSum(root->right,k,sum); } int sum(Node* root, int k) { // Your code here int sum = 0; getSum(root , k , sum); return sum; }

[@chanchalsoni7535]

ans of 3rd que void solve(Node* root,int &sum,int &k){ if(!root)return; solve(root->left,sum,k); if(k!=0){sum=sum+root->data; k=k-1; solve(root->right,sum,k); } } int sum(Node* root, int k) { // Your code here int sum=0; solve(root,sum,k); return sum; }

[@Sanataniblood-mb5pu]

Awesome content bhaiya ❤️

[@dayashankarlakhotia4943]

class Solution { int ans=Integer.MAX_VALUE;prev=-1; public int minDiffInBST(TreeNode root){ if(root.left!=null) minDiffInBST(root.left); if(prev!=-1) ans=Math.min(ans,root.val-prev); prev=root.val; if(root.right!=null) minDiffInBST(root.right); return ans; } }

[@universalmeditationmusic9299]

nice

[@vishalgupta-gi7kr]

Sir please make video on web developer

[@GamingArenaaa]

printf("all clear"); printf("🔥🔥🔥🔥");

[@allinonemoviesyt]

use cout

[@push_3872]

To check the BST to checking the right > node and left < Node for each tree is not correct

[@deep_singh01]

Day 165/180 👍👍

[@dhirajpatil7453]

This course videos will be remain free forever or you will make it private because I have just started to watch this playlist sir..

[@sumitvishwakarma1075]

bhai hamesha free rahega aap kabhi bhi dekh sakte ho

[@amarnathtripathy4350]

array to bst question changed ?

[@ahmadhassan90590]

yes you can only do with tree method. now array method is not accepting

[@crazydude8333]

Bhaiya thought process kaise develop hoga DSA me bohot struggle hota hai??

[@shrishtdev9961]

bhaiya ji namste❤❤❤

[@Pallab01]

4: Kth largest element in BST void kLargest(Node *root,int &large,int &k) { //RNL if(!root) return ; // right side kLargest(root->right,large,k); if(k>0) { large=root->data; k--; } else return ; //left side kLargest(root->left,large,k); } int kthLargest(Node *root, int K) { //Your code here int large; kLargest(root,large,K); return large; }

[@shubhamkumarjha9192]

Good morning bhaiya ji💖, kaise hai tabiyat bhaiya ji aapki ab..?

[@CoderArmy9]

Good Morning bhai❤, Ab sahi hai bhai

[@anshyoutuber1608]

Bhaiya is Sunday live aayiye please

[@SayedHasinoor]

Bhaiya BST to Double linkedlist & vice versa problem karo

[@Eng_wallah]

Har har Mahadev 🙏🏻🙏🏻

[@RajputArjun-ss4wj]

Bhaiya java me bhi ek dsa course banaye

[@allinonemoviesyt]

2:03 nhi bhaiya

[@itshirdeshk]

Day 165 ✅🔥

[@ankushladani496]

Bhaiya 1st question ka code part khudse likha...😊

[@YashSaini007]

Bhaiya ab aap thik ho.

[@CoderArmy9]

Haan bhai❤

[@YashSaini007]

Day 164/180 #180DaysOfCode

[@chaolobi7698]

GFG kth largest : - void inorder(Node *root, int &ans, int &k) { if (!root) { return; } if (k right, ans, k); if (k > 0) { ans = root->data; k--; } inorder(root->left, ans, k); } int kthLargest(Node *root, int K) { // Your code here int ans; inorder(root, ans, K); return ans; }

[@Karamjeet_11]

❤

[@joychakraborty4367]

like done

[@YASHYADAV-f9i]

bhaiya maja a gaya

[@averagecoder7]

❤❤❤❤

[@Ramgangakumar]

bhaiya app kon se year m ho 😂😂😂😂

[@gajendrasinghdhaked]

🤩

[@DelightfulTreeSwing-qo6tm]

Nhi 2:00 logic shi nhi h

[@akashkumawat-xk1eg]

Logic galat h ki har node ko check kare left small and right bigger ho

[@sabiruddinkhan496]

T.C:O(N); S.C:O(H); H->height of the BST; class Solution { public: Node*converttobst(vector&nums,Node*root,int s,int e){ if(s>e) return NULL; int mid=s+(e-s)/2; root=new Node(nums[mid]); root->left=converttobst(nums,root->left,s,mid-1); root->right=converttobst(nums,root->right,mid+1,e); return root; } Node* sortedArrayToBST(vector& nums) { Node*root=NULL; return converttobst(nums,root,0,nums.size()-1); } };

[@DelightfulTreeSwing-qo6tm]

Nhi 2:00 logic shi nhi h

[@utkarshrathore2990]

😍😍