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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.

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** Link for the Problem** – Populating Next Right Pointers in Each Node

Populating Next Right Pointers in Each Node– LeetCode Problem

**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:**

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:[]

**Constraints:**

- The number of nodes in the tree is in the range
`[0, 2`

.^{12}- 1] `-1000 <= Node.val <= 1000`

**Follow-up:**

- 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 { public: Node* connect(Node* root) { if (!root) return nullptr; connectTwoNodes(root->left, root->right); return root; } private: void connectTwoNodes(Node* p, Node* q) { if (!p) return; 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: return 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