Binary Tree Problems: Mirror Tree | Two Nodes are Cousins | Check for Balanced Tree | Spiral Form

[@CoderArmy9]

Like Target 1k✅

[@DelightfulTreeSwing-qo6tm]

Bhaiya March main 100k

[@joychakraborty4367]

Done bhaiya

[@vikaspandit-f3h]

ho jaayega sir aag jaruur lagega

[@Gurmeet-z8u]

Day 152/180 ki video kha h?

[@vinaymaurya4895]

Yar views 3k aur likes sirf 200 itna quality content mil rha h free me, ab kya chahiye bhai log? Har kisi ne laptop me pdha diya h sare courses me lekin sirf rohit bhaiya ne hi itna quality content ko board pe cover kr rhe hn, kitna efforts lgate hn, koi faltu bkwas nhi koi promotion nhi na hi koi faltu ka excuse ki maine job aap logon ko pdhane k liye quit kr di, itna high quality content de rhe hn thoda support krne me kya dikkat h,plz bhai support kro yr plz humble request sbse 🙏🙏🙏

[@RahulGupta-wv7lv]

Bhai ye fake id rohit bhai ke he hai kya

[@rasheedahmad5611]

your teaching style is just amazing. soon you will touch 1M subscriber.

[@DelightfulTreeSwing-qo6tm]

Bhaiya maine 4 problems khud se solve kiya tha except for spiral transversal ….. bhaiya apke videos se mujhe bhut kuch sikhne ko mila hai … dry run and thinking process … thanks for teaching us .

[@nidhiraj9686]

Bhaiya 2 question khud se hue. Boht acha lecture tha ye. Boht kuch sikhne ko mila ❤❤

[@Aniruddha_Mukherjee]

Bhaiya 1 problem khudse solve kar liya tha , aur 2nd problem ka algo khud se soch liya tha lekin code nhi ho paya. Aaj ke lecture se bohot kuch seekha , aur logic building bhi improve ho raha hain. 😊

[@neelmishra5304]

51:34 ans- if(valid) conditional statement was written before so that whenever the valid becomes zero we dont need to traverse the whole tree and it return from the node where valid becomes zero.

[@akashkewat1219]

Aaj Leetcode POTD mei bhi same question tha bhaiya. Code kafi accche hai tree ke.

[@chanchalsoni7535]

highly underrated channel😐 why dont people like the videos....i feel so bad .....

[@fachoyt9686]

I feel so much more confident in Binary Tree after watching your videos. Thanks for being such a great resource!

[@zafrul_islam]

bhaiya khud se ek bhi problem solve nhi hui but maja aaya sikh kr 😊😊

[@rrr9848]

Congratulations bhaiya 🎉

[@asad_iliyas]

51:00 : if(valid==0) => It means unbalanced node is found and we will directly come out of function by returning -1 , and our valid would have become 0 and we will return it in our main function.

[@kamleshregar5880]

very good lectures. keep doing it for us. thank you bro for your hardwork

[@heetpatel3037]

Aapka jitna shukriya da karu utna kam hai

[@alltech_info]

Shuru majboori me kiye the ab maja aa rha hai😇😂

[@devendrasharma8750]

bhaiya did level order traversal in spiral by my self. thanku bhaiya❤‍🩹

[@Noobiee_FPS]

sab kuch kr liya bhaiya sare questions kr liya spiral wale code mai confuse ho gya tha par dry run kra toh sab smjh mai aa gya tha first three question approach ka idea lga liya tha bss code nhi kra tha dono question video dekh ke ek cheez smjha aaya tree ko traverse krna aana chaiye tabhi question hoge

[@heetpatel3037]

Jai shree Ram ❤️ Bohot bohot dhanyawad aapka bhaiya

[@Sanataniblood-mb5pu]

Content outstanding and bhaiya is also outstanding 😂🙏🏻

[@RohitKumar-gj2px]

Thanku so much for best quality content . Your style of teaching is amazing and i am enjoying your videoes . LOVE YOU BROTHER ❤❤

[@ankushladani496]

Bhaiya Khali second wali if statement nahi kar paya Baki pura solution khudse implement kar liya tha...😊❤

[@anupammishra6514]

1:44:54 maja aa gya bhaiya ❤❤......1 questions solve hua hamse 🥲🥲

[@um6819]

UPDATED CODE if test case not passed in GFG for TC no. 1075 (see the previous line before the last line) everything will be same bool isParent(Node *root, int a, int b){ if(root==NULL) return 0; if(root->left && root->right){ if(root->left->data == a && root->right->data == b) return 1; if(root->left->data == b && root->right->data == a) return 1; } return (isParent(root->left, a, b) || isParent(root->right, a, b)); } bool isCousins(Node *root, int a, int b) { queue q; q.push(root); int level = 0; int l1 = -1, l2 = -1; while(!q.empty()){ int n = q.size(); while(n--){ Node* temp = q.front(); q.pop(); if(temp->data==a) l1 = level; if(temp->data==b) l2 = level; if(temp->left) q.push(temp->left); if(temp->right) q.push(temp->right); } level++; if(l1!=l2) return 0; if(l1!=-1) break; } if(l1==-1 && l2==-1) return 0; //updated line, if a and b is not present anywhere in the tree return !isParent(root, a, b);

[@CoderArmy9]

Good Job bhai, they have updated the TC.

[@haardikmadaan1644]

thanks bro....i was so confused in this

[@nownow1025]

it still fails t.c 219

[@JatinKansal-z5b]

Still not working plz help

[@nownow1025]

Great explanation bhaiya. Thank you so much ♥

[@ankushladani496]

Sab kuch chamak gaya Bhaiya...❤🎉

[@rishavdobriyal8201]

amazing explaination sir

[@akashkumawat-xk1eg]

Maza aa gaya❤❤ I solved all problems by myself😀😀

[@Vardanaggarwal]

thanks bhaiya amazing series leking ek bhi problem khud se nhi huyi.

[@GOATSOFFOOTBALL]

Thodasa brainstorming tha session but loved it,best explanation

[@amanprajapati6301]

thank u bhaiya for providing quality content dil se

[@sunnyrai4329]

Congratulations❤ almost 3 que solved by self

[@AnordinaryMan007]

Best course ❤

[@Md.FarhanMasud-u8x]

// Spiral Print Binary tree using deque vector findSpiral(Node *root) { int n; vector ans; deque q; q.push_back(root); while (q.size()) { // print Right to left & Reverse left to right n = q.size(); while (n--) { Node *temp = q.back(); q.pop_back(); ans.push_back(temp->data); if (temp->right) q.push_front(temp->right); if (temp->left) q.push_front(temp->left); } // Print left to right and Reverse right to left n = q.size(); while (n--) { Node *temp = q.front(); q.pop_front(); ans.push_back(temp->data); if (temp->left) q.push_back(temp->left); if (temp->right) q.push_back(temp->right); } } return ans; }

[@rishusachan2618]

Thanku Bhaiya, ekdum accha say chmak gya 😍😇

[@ayushpanwar1671]

bhaiya sab samaj agya... one day you will be recognized

[@rocketgamerzz.7577]

Problem 4 can be done using a single deque instead of two stacks for odd level push at back and take from front and for even level vice versa here's the code for your reference deque q; q.push_back(root); vector ans; while(!q.empty()) { int n = q.size(); while(n--) { //odd level k liye Node *temp = q.back(); q.pop_back(); ans.push_back(temp->data); if(temp->right) { q.push_front(temp->right); } if(temp->left) { q.push_front(temp->left); } } n = q.size(); // even level k liye while(n--) { Node *temp=q.front(); ans.push_back(temp->data); q.pop_front(); if(temp->left) { q.push_back(temp->left); } if(temp->right) { q.push_back(temp->right); } } }

[@kumarashutosh-p3n]

class Solution { public: //Function to check if two trees are identical. bool isIdentical(Node *r1, Node *r2) { vectora; vectorb; queuep; queueq; Node* temp; Node* dummy; p.push(r1); q.push(r2); while(!p.empty()){ temp=p.front(); p.pop(); a.push_back(temp->data); if(temp->left) p.push(temp->left); if(!temp->left) a.push_back(0); if(!temp->right) a.push_back(0); if(temp->right) p.push(temp->right); } while(!q.empty()){ dummy=q.front(); q.pop(); b.push_back(dummy->data); if(dummy->left) q.push(dummy->left); if(!dummy->left) b.push_back(0); if(!dummy->right) b.push_back(0); if(dummy->right) q.push(dummy->right); } if(a==b) return true; else return false; } };

[@KUNALGAUTAM-bf1yy]

iss tarike se likhna hoga code aage khain bhi? like exam intreview

[@amybross]

thank you brother for your great effort

[@5harshitshrivastava]

superb quality content🙌

[@Eng_wallah]

Radhe Radhe bhaiya 🙏🏻🙏🏻

[@neetuchaudhary7884]

thannnkkkkk youuuuuuu so much bhaiya🤗🤗

[@ujjwal_7531]

i did the level order traversal in spiral question using queue vector findSpiral(Node *root) { //Your code here queueq; vectorans; q.push(root); int rand=1; while(!q.empty()){ int size=q.size(); vectortem; for(int i=0;idata); if(temp->left) q.push(temp->left); if(temp->right) q.push(temp->right); } if(rand%2==1) for(int i=size-1;i>=0;--i) ans.push_back(tem[i]); else for(int i=0;i

[@vijaysingh-n6p9j]

Bahut badiya tha bhaiya

[@study-yd6es]

Amazing Videos keep it up brother

[@Harshmishra0007]

Last question Two Nodes are cousin int level(Node *root,int i,int a){ if(root==NULL) return 0; if(root->data==a){ return i; } return max(level(root->left,i+1,a),level(root->right,i+1,a)); } bool f(Node * root,int a,int b,int m,int i){ if(root==NULL) return true; if(root->left && root->right){ if(root->left->data==a && root->right->data==b) return false; else if(root->left->data==b && root->right->data==a) return false; } if(i==m) return true; return (f(root->left,a,b,m,i+1) && f(root->right,a,b,m,i+1)); } bool isCousins(Node *root, int a, int b) { if(!root->left && !root->right) return false; int m=level(root,0,a); int n=level(root,0,b); if(m!=n) return false; else return f(root,a,b,n,0); }

[@soumyadeepghosh4428]

// Approach 2 for Problem - 3 (Balanced Binary Tree) class Solution{ public: int height(Node* root) { if(!root) return 0; int leftHeight = height(root->left); int rightHeight = height(root->right); return 1 + max(leftHeight, rightHeight); } bool isBalanced(Node *root) { if(!root) return 1; if(abs(height(root->left)-height(root->right)) > 1) return 0; return isBalanced(root->left) and isBalanced(root->right); } };

[@xlr8-Ujjwal]

For those whose program is not passing after test case 1075. Simply add the below line to resolve the edge case ⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇ if(!root ->left && !root ->right) return 0;

[@dipakmali9817]

Day 154/180 ✅

[@Aniruddha_Mukherjee]

Good morning Bhaiya ji😍

[@vaishnavigoswami5707]

problem 5: to find level of a and b using recursion h =0; and l1=-1;l2=-1;intially void level(Node* root,int a,int b,int h,int&l1,int&l2) { if(root==NULL) return; h++; if(root->left) { if(root->left->data==a) l1=h; if(root->left->data==b) l2 =h; } if(root->right) { if(root->right->data==a) l1=h; if(root->right->data==b) l2 =h; } level(root->left,a,b,h+1,l1,l2); level(root->right,a,b,h+1,l1,l2); }

[@ChandanKumar-kt8ku]

Congratulations

[@YashSaini007]

Bhaiya Radhe Radhe 🙏

[@pankajdubey1123]

Level order Traversal in spiral form : vector findSpiral(Node *root) { //Your code here deque dq; dq.push_back(root); vector ans; int flag=0; int n; Node *temp; while(!dq.empty()) { if(flag%2==0) { n=dq.size(); while(n--) { temp=dq.back(); dq.pop_back(); ans.push_back(temp->data); if(temp->right) dq.push_front(temp->right); if(temp->left) dq.push_front(temp->left); } } else { n=dq.size(); while(n--) { temp=dq.front(); dq.pop_front(); ans.push_back(temp->data); if(temp->left) dq.push_back(temp->left); if(temp->right) dq.push_back(temp->right); } } flag++; } return ans; }

[@vivektechtuber204]

Good Morning Bhaiya 🙏

[@ankushladani496]

Thanks bhaiya...❤🎉

[@mohit6215]

100k laga do bahiyo

[@SanatanaDharmaDaivagnya]

jai ho bhaiyya

[@JayPrajapati-lj8pw]

Best content😊😊❤

[@aahilkhan4280]

2 questions khud se kr liye bhaiyya ,remaining 3 problems ka algo bhi likh liya tha ,but implement nhi kr paya.

[@aadityakumarjha-ju3xg]

Good morning bhaiya

[@punitsharma584]

me in this last question when level came in my mind be like ::level sabke niklenge pr niklenge uske jo khada rhega

[@deep_singh01]

Day 154/180 👍👍

[@asad_iliyas]

4/5 solved

[@itshirdeshk]

Day 154 ✅🔥

[@allinonemoviesyt]

1:44:14 Homework bool parents(Node *root, int a, int b){ if(!root) return false; if(root->left && root->right){ if(root->left->data == a && root->right->data == b){ return true; } if(root->left->data == b && root->right->data == a){ return true; } } return parents(root->left,a,b) || parents(root->right,a,b); } void findLevel(Node *root, int a, int b, int level, int &l1 , int&l2){ if(!root){ return; } if(root->left){ if(root->left->data == a){ l1 = level+1; } if(root->left->data == b){ l2 = level+1; } } if(root->right){ if(root->right->data == a){ l1 = level+1; } if(root->right->data == b){ l2 = level+1; } } if(l1 == -1 || l2 == -1) findLevel(root->left,a,b,level+1,l1,l2); if(l1 == -1 || l2 == -1) findLevel(root->right,a,b,level+1,l1,l2); } bool isCousins(Node *root, int a, int b) { int l1 = -1, l2 = -1; findLevel(root,a,b,1,l1,l2); if(l1 != l2) return false; if(parents(root,a,b)){ return false; } return true; }

[@utkarshpandey1394]

nice

[@PrathamSinghhAir]

51:58 pe sir ka " dum dum dum dum " 😂

[@vijaysingh-n6p9j]

bhaiya last vala question bahut hard tha

[@ankitchoudhary2914]

Chill scene❤❤❤

[@pradeepraghu1see]

till end .

[@YASHYADAV-f9i]

bhaiya maja aagya

[@akshaythakur4231]

Wow ❤

[@fakharabbas9738]

Love U Coder Army❤

[@YashSaini007]

Day 154/180 #180DaysOfCode

[@abhishekgupta7709]

Bhaiya chamka. One request put the Coder army DSA sheet on its website .

[@Abhishekkumar-gr5vr]

🔥🔥

[@ankitshrestha6481]

// Time O(N) and Space O(N) // Another approach bool isNodes_areCousin(Node* root, int first, int second) { queue list; list.push(root); list.push(nullptr); int currentLevel = 0; int firstLevel = 0; int secondLevel = 0; int parent1 = -1; int parent2 = -1; Node* temp; while (!list.empty()) { temp = list.front(); list.pop(); if (temp == nullptr) { if (list.size() == 0) break; currentLevel++; list.push(nullptr); } else { if (temp->left) { if (temp->left->data == first) { parent1 = temp->data; firstLevel = currentLevel + 1; } if (temp->left->data == second) { parent2 = temp->data; secondLevel = currentLevel + 1; } list.push(temp->left); } if (temp->right) { if (temp->right->data == first) { parent1 = temp->data; firstLevel = currentLevel + 1; } if (temp->right->data == second) { parent2 = temp->data; secondLevel = currentLevel + 1; } list.push(temp->right); } } } return (parent1 != parent2) && (firstLevel == secondLevel); }

[@kuldeepparmar9861]

chamak gaya last tk

[@hiteshsharma5034]

Bhaiya, Here is my code for Vertical Traversal using pair vector concept, sorting and recursion... class Solution { public: void find(TreeNode* root, int pos, int level, vector &v){ if(root==NULL){ return; } v.push_back(make_pair(pos, make_pair(level, root->val))); find(root->left,pos-1, level+1,v); find(root->right, pos+1, level+1,v); } vector verticalTraversal(TreeNode* root) { vector v; vector ans; if(root==NULL){ return ans; } find(root,0,0,v); sort(v.begin(),v.end()); int prev=v[0].first; vector temp; for(auto x: v){ if(x.first==prev){ temp.push_back(x.second.second); } else{ prev=x.first; ans.push_back(temp); temp.clear(); temp.push_back(x.second.second); } } ans.push_back(temp); return ans; } };

[@amorphous8826]

👍

[@Aryan-wl7mc]

0:28 Google mai job pakki hai jisne like kiya 😂😂😂

[@yashpaunikar671]

bhaiya leetcode ke bhi link attach kr dijiye

[@meetdobariya8777]

same code likha cousin ka lekin 1075 pe differnce a raha he

[@xlr8-Ujjwal]

Add this line if(!root ->left && !root ->right) return 0;

[@fakharabbas9738]

Love U Coder Army

[@IR20025]

Lec - 104

[@pranavmittal9619]

2/5 questions ho gaye lekin vo bhi pehle kar rakhe the

[@swarupdwivedy5733]

How to prepare DSA with GATE?

[@ankitchoudhary2914]

done todays classs

[@vishnusingh2300]

29:00

[@rishuagarwal1523]

leet code 5th number question testcase 1075 is giving wrong output with same code can anybody help

[@farhansadik7236]

in the spiral traversal i got TLE in leetcode,i cannot figure it out . anyone ?

[@letsk9188]

i reach to last

[@gateavi]

#1

[@ShaikSharma]

sir,i have my google interview can i crack if i watch all of your playlist

[@anupammishra6514]

❤❤

[@vivek7187]

🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥

[@vishnusingh2300]

51.21

[@prashantsingh3474]

Level order traversal in spiral form solution of this question using deque... vector findSpiral(Node *root) { dequeq; vectorans; int count=0; q.push_back(root); while(q.empty()!=1){ int qsize=q.size(); for(int i=0;idata); if(temp->right!=NULL){ q.push_front(temp->right); } if(temp->left!=NULL){ q.push_front(temp->left); } } else{ //if count is odd means in tree we move right to left Node* temp=q.front(); q.pop_front(); ans.push_back(temp->data); if(temp->left!=NULL){ q.push_back(temp->left); } if(temp->right!=NULL){ q.push_back(temp->right); } } } count++; } return ans; } If any anyone like my solution then plz like my comment...

[@haardikmadaan1644]

1/5