Populating Next Right Pointers in Each Node | Tree Data Structure playlist C++ | Hello World

This is the video under the series of DATA STRUCTURE & ALGORITHM in a TREE Playlist. We are going to understand How to Populating Next Right Pointers in Each Node.
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► 116. Populating Next Right Pointers in Each Node
You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
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[@girikgarg8]

There's a follow-up which asks us to do this question using constant extra space (recursive stack space is not considered as extra space). How would you do it then?

[@surjeetsingh433]

Keep growing brother 💙

[@HelloWorldbyprince]

We will

[@shaantyagi2187]

Aa gya samajh bhaiya, time laga magar aa gya ...ek baar fir se level order traversal revise kiya tab samajh gye!

[@HelloWorldbyprince]

bass aise hi revision karte rho samjh me aa jayega chiz

[@ujjwalsharmaiiitunacse3884]

Great content sir !! keep up the good work

[@vanivaishnavi3916]

Well well...you arr best

[@HelloWorldbyprince]

thanks a lot vani

[@RYFINTECH]

C++ || Easy to understand solution class Solution { public: Node* connect(Node* root) { if(root == NULL) return root; queue q; q.push(root); while(!q.empty()){ int size = q.size(); for(int i=0;inext = q.front(); // last element na ho, 0 based h if(temp ->left != NULL) q.push(temp->left); if(temp->right != NULL) q.push(temp->right); } } return root; } };

[@HelloWorldbyprince]

Amazing yaar ✅

[@thegreekgoat98]

I DID THIS USING SIMPLE DFS:: Node* connect(Node* root) { Node* black=root; while(black!=NULL && black->left!=NULL) { Node* n=black; while(true) { n->left->next=n->right; if(n->next==NULL) break; n->right->next=n->next->left; n=n->next; } black=black->left; } return root; }

[@MANISHKUMAR-vn1yh]

MOJ krdi bhaiyya

[@HelloWorldbyprince]

Yeah 😂😂

[@harshsrivastava2075]

class Solution { public: Node* connect(Node* root) { if (!root) { return root; } queue q; q.push(root); while (!q.empty()) { int sizee = q.size(); for (int i = 0; i < sizee; i++) { Node* current = q.front(); q.pop(); if (i != sizee - 1) { current->next = q.front(); } if (current->left) { q.push(current->left); } if (current->right) { q.push(current->right); } } } return root; } }; more optimized solution. Wrote it without seeing the solution!

[@gamerversez5372]

My Code in C++ Node* connect(Node* root) { if(!root) return NULL; queue q; q.push(root); while(!q.empty()) { int size = q.size(); vector x; while(size!=0) { if(size==0) break; else if(size==1) { auto i=q.front(); q.pop(); if(i->left) q.push(i->left); if(i->right) q.push(i->right); break; } else if(size>1) { auto i=q.front(); q.pop(); if(i->left) q.push(i->left); if(i->right) q.push(i->right); auto nex = q.front(); i->next=nex; size-=1; } } } return root; }

[@HelloWorldbyprince]

good work

[@shubhambhatt2704]

Why is this question marked medium though it is easy?

[@HelloWorldbyprince]

idk

[@pritishpattnaik4674]

I guess we can use level order traversal here

[@HelloWorldbyprince]

yo

[@cr7johnChan]

if(root==NULL)return root; queueq; q.push(root); Node*temp; while(q.empty()!=1){ int s =q.size(); while(s){ temp=q.front(); q.pop(); if(s==1)temp->next==NULL; else{ temp->next=q.front(); } if(temp->left!=NULL)q.push(temp->left); if(temp->right!=NULL)q.push(temp->right); s--; } } return root; }

[@HelloWorldbyprince]

Ekdum jhakaas

[@harshalgarg1149]

This solution is better. Check this out. Node* connect(Node* root) { if(root==NULL) return root; queue q; q.push(root); while(!q.empty()) { int size = q.size(); while(size--) { Node* top = q.front(); q.pop(); if(top->left) q.push(top->left); if(top->right) q.push(top->right); if(size!=0) top->next = q.front(); } } return root; }