Last updated on October 5th, 2024 at 09:23 pm

Here, We see ** Populating Next Right Pointers in Each Node LeetCode Solution**. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

*List of all LeetCode Solution*

*List of all LeetCode Solution*

## Topics

Array, Hash Table, Two-Pointers

## Companies

## Level of Question

Medium

**Populating Next Right Pointers in Each Node LeetCode Solution**

## Table of Contents

**Problem Statement**

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:**

**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:** []

**1. Populating Next Right Pointers in Each Node LeetCode Solution C++**

class Solution { public: Node* connect(Node* root) { if(root == NULL) return NULL; if(root->left != NULL) root->left->next = root->right; if(root->right != NULL && root->next != NULL) root->right->next = root->next->left; connect(root->left); connect(root->right); return root; } };

**2. Populating Next Right Pointers in Each Node Solution Java**

class Solution { public Node connect(Node root) { if(root == null) return null; if(root.left != null) root.left.next = root.right; if(root.right != null && root.next != null) root.right.next = root.next.left; connect(root.left); connect(root.right); return root; } }

**3. Populating Next Right Pointers in Each Node Solution JavaScript**

var connect = function(root) { if (root == null) return root; let queue = [root]; while(queue.length!=0) { let next = []; while(queue.length!=0) { let node = queue.shift(); node.next = queue[0]||null; if (node.left!=null) { next.push(node.left); next.push(node.right); } } queue = next; } return root; };

**4. Populating Next Right Pointers in Each Node Solution Python**

class Solution(object): def connect(self, root): if not root: return c1, c2 = root.left, root.right while c1 and c2: c1.next = c2 c1, c2 = c1.right, c2.left self.connect(root.left) self.connect(root.right) return root