Last updated on October 5th, 2024 at 05:47 pm

Here, We see ** Populating Next Right Pointers in Each Node II 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

Depth-First Search, Tree

## Companies

Bloomberg, Facebook, Microsoft

## Level of Question

Tree

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

## Table of Contents

**Problem Statement**

Given a binary treestruct 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,null,7] **Output:** [1,#,2,3,#,4,5,7,#] **Explanation: **Given the above 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 II LeetCode Solution C++**

class Solution { public: Node* connect(Node* root) { Node *currParent = root, *baseChild, *currChild, *nextChild; while (currParent) { while (currParent->next && !currParent->left && !currParent->right) currParent = currParent->next; currChild = baseChild = currParent->left ? currParent->left : currParent->right; while (currChild) { if (currParent->right && currChild != currParent->right) nextChild = currParent->right; else { currParent = currParent->next; while (currParent && !currParent->left && !currParent->right) currParent = currParent->next; nextChild = currParent ? currParent->left ? currParent->left : currParent->right : currParent; } currChild->next = nextChild; currChild = nextChild; } currParent = baseChild; } return root; } };

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

class Solution { public Node connect(Node root) { if (root == null) { return root; } Queue<Node> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { int levelCount = queue.size(); Node prev = null; for (int i = 0; i < levelCount; i++) { Node curNode = queue.poll(); if (prev != null) { prev.next = curNode; } prev = curNode; if (curNode.left != null) { queue.add(curNode.left); } if (curNode.right != null) { queue.add(curNode.right); } } } return root; } }

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

var connect = function(root) { if (!root) return root; let queue = [root]; let tempQueue = []; while(queue.length){ let curr = queue.splice(0, 1)[0]; let {left, right} = curr; if (left) tempQueue.push(left); if (right) tempQueue.push(right); if (queue.length === 0){ curr.next = null; queue = tempQueue; tempQueue = []; }else{ curr.next = queue[0]; } } return root; };

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

class Solution(object): def connect(self, root): if not root: return None q = deque() q.append(root) dummy=Node(-999) while q: length=len(q) prev=dummy for _ in range(length): popped=q.popleft() if popped.left: q.append(popped.left) prev.next=popped.left prev=prev.next if popped.right: q.append(popped.right) prev.next=popped.right prev=prev.next return root