Last updated on July 19th, 2024 at 11:32 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

Topics

Array, Hash Table, Two-Pointers

Companies

LinkedIn

Level of Question

Medium

Populating Next Right Pointers in Each Node LeetCode Solution

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