Ppulaing Nx igh Pins in ah Nd Ld Pgamming Sluins | Ld Pblm Sluins in ++, Java, & Pyhn []

LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++

Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Populating Next Right Pointers in Each Node in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

Use “Ctrl+F” To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Option to Get Any Random Questions Answer.

About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
  • Arrays
  • Strings
  • Hash Table
  • Dynamic Programming
  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the Problem Populating Next Right Pointers in Each Node

Populating Next Right Pointers in Each Node– LeetCode Problem


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.

Example 1:

116 sample
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.

Example 2:

Input: root = []
Output: []


  • The number of nodes in the tree is in the range [0, 212 - 1].
  • -1000 <= Node.val <= 1000


  • You may only use constant extra space.
  • The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.
Populating Next Right Pointers in Each Node– LeetCode Solutions
Populating Next Right Pointers in Each Node Solution in C++:
class Solution {
  Node* connect(Node* root) {
    if (!root)
      return nullptr;
    connectTwoNodes(root->left, root->right);
    return root;

  void connectTwoNodes(Node* p, Node* q) {
    if (!p)
    p->next = q;
    connectTwoNodes(p->left, p->right);
    connectTwoNodes(q->left, q->right);
    connectTwoNodes(p->right, q->left);
Populating Next Right Pointers in Each Node Solution in Java:
class Solution {
  public Node connect(Node root) {
    Node node = root; // the node just above current needling

    while (node != null && node.left != null) {
      Node dummy = new Node(); // dummy node before needling
      // needle children of node
      for (Node needle = dummy; node != null; node = node.next) {
        needle.next = node.left;
        needle = needle.next;
        needle.next = node.right;
        needle = needle.next;
      node = dummy.next; // move node to the next level

    return root;
Populating Next Right Pointers in Each Node Solution in Python:
class Solution:
  def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
    if not root:
      return None

    def connectTwoNodes(p, q) -> None:
      if not p:
      p.next = q
      connectTwoNodes(p.left, p.right)
      connectTwoNodes(q.left, q.right)
      connectTwoNodes(p.right, q.left)

    connectTwoNodes(root.left, root.right)
    return root

Leave a Comment

Scroll to Top