What is the time complexity of Populating Next Right Pointers in Each Node?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Populating Next Right Pointers in Each Node?

This problem uses Linked List, Tree, Depth-First Search, Breadth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Populating Next Right Pointers in Each Node?

Insight 1: The perfect binary tree structure allows for a constant space solution by leveraging existing links to traverse the tree level by level. Insight 2: The solution can avoid explicit level-order traversal (BFS) using the `next` pointers to move horizontally across each level.

Which companies ask LeetCode 116. Populating Next Right Pointers in Each Node in interviews?

This problem is frequently asked at Citadel, Flipkart, Google, Oracle, Salesforce and 3 more companies.

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Populating Next Right Pointers in Each Node - LeetCode 116 Solution

Populating Next Right Pointers in Each Node - Complete Solution Guide

Populating Next Right Pointers in Each Node is LeetCode problem 116, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 p

Detailed Explanation

The problem asks us to connect all the nodes at the same level of a given perfect binary tree using the `next` pointer. A perfect binary tree is one where all internal nodes have two children and all leaves are at the same level. Initially, all `next` pointers are null. The goal is to populate each node's `next` pointer to point to the node immediately to its right at the same level. If a node is the rightmost node at its level, its `next` pointer should remain null.

Solution Approach

The provided solution uses a level-order traversal approach, but it does so without using a queue or any other data structure for explicit level tracking. It iterates through the tree level by level, starting from the root. For each level, it traverses all nodes and connects the `next` pointers accordingly. The outer loop moves from one level to the next (going down the leftmost node), and the inner loop connects nodes within the same level.

Step-by-Step Algorithm

Step 1: Check if the root is null. If it is, return null.
Step 2: Initialize `leftmost` to the root. `leftmost` represents the leftmost node of the current level.
Step 3: While `leftmost` has a left child (meaning there are more levels to process):
Step 4: Initialize `head` to `leftmost`. `head` represents the current node we are processing on the current level.
Step 5: While `head` is not null (meaning we haven't reached the end of the current level):
Step 6: Connect the left child's `next` pointer to the right child: `head.left.next = head.right;`.
Step 7: If `head` has a `next` node (meaning it's not the rightmost node on the level):
Step 8: Connect the right child's `next` pointer to the left child of the `next` node: `head.right.next = head.next.left;`.
Step 9: Move to the next node on the level: `head = head.next;`.
Step 10: Move down to the next level by setting `leftmost` to its left child: `leftmost = leftmost.left;`.
Step 11: Return the root.

Key Insights

Insight 1: The perfect binary tree structure allows for a constant space solution by leveraging existing links to traverse the tree level by level.
Insight 2: The solution can avoid explicit level-order traversal (BFS) using the `next` pointers to move horizontally across each level.
Insight 3: Recognizing the pattern of connecting `left` to `right` child within the same node and `right` child to `left` child of the `next` node.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Tree, Depth-First Search, Breadth-First Search, Binary Tree.

Companies

Asked at: Citadel, Flipkart, Google, Oracle, Salesforce, ServiceNow, Snowflake, Walmart Labs.

Populating Next Right Pointers in Each Node | LeetCode 116

Medium

Linked List Tree Depth-First Search Breadth-First Search Binary Tree

Time: O(n)Space: O(1)

Solution Code

Available in:

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return null;
        }

        Node leftmost = root;

        while (leftmost.left != null) {
            Node head = leftmost;
            while (head != null) {
                head.left.next = head.right;

                if (head.next != null) {
                    head.right.next = head.next.left;
                }

                head = head.next;
            }

            leftmost = leftmost.left;
        }

        return root;
    }
}

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Problem Info

Description

Problem 116: Populating Next Right Pointers in Each Node - Detailed explanation and solution approaches.

Topics

Linked List Tree Depth-First Search Breadth-First Search Binary Tree

Companies

CitadelFlipkartGoogleOracleSalesforceServiceNowSnowflakeWalmart Labs

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:116

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Check if the root is null. If it is, return null.
2
Step 2: Initialize `leftmost` to the root. `leftmost` represents the leftmost node of the current level.
3
Step 3: While `leftmost` has a left child (meaning there are more levels to process):
4
Step 4: Initialize `head` to `leftmost`. `head` represents the current node we are processing on the current level.
5
Step 5: While `head` is not null (meaning we haven't reached the end of the current level):
6
Step 6: Connect the left child's `next` pointer to the right child: `head.left.next = head.right;`.
7
Step 7: If `head` has a `next` node (meaning it's not the rightmost node on the level):
8
Step 8: Connect the right child's `next` pointer to the left child of the `next` node: `head.right.next = head.next.left;`.
9
Step 9: Move to the next node on the level: `head = head.next;`.
10
Step 10: Move down to the next level by setting `leftmost` to its left child: `leftmost = leftmost.left;`.
11
Step 11: Return the root.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because each node is visited only once during the level-order traversal using the `next` pointers. The space complexity is O(1) because the solution operates in-place and doesn't use any additional data structures that scale with the input size, satisfying the constant extra space constraint.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS) with Queue

Use a queue to perform a standard level-order traversal. At each level, iterate through the nodes in the queue and connect their `next` pointers.

Time: O(n)Space: O(w), where w is the maximum width of the tree. In a perfect binary tree, w = n/2 at the last level.

Trade-offs: This approach is easier to understand for beginners familiar with BFS, but it uses O(w) space, which can be significant for large perfect binary trees, failing the O(1) space requirement. The primary advantage of the constant space approach is its memory efficiency.

Recursive Approach

A recursive approach can be implemented, though it can be less efficient in some languages due to function call overhead. Recursively connect the left and right subtrees, and handle the connection between the rightmost node of the left subtree and the leftmost node of the right subtree. Although the problem states recursion is fine, it's crucial to acknowledge the implicit call stack's space usage.

Time: O(n)Space: O(log n) in the best/average case (balanced tree) due to the call stack, and O(n) in the worst case (skewed tree). Although the problem states to assume the implicit stack space does not count as extra space, it is important to remember its effects.

Trade-offs: This approach may be more intuitive for some, but it comes with the overhead of the call stack, which could affect performance and violate the constant space requirement in worst case scenarios (especially on severely unbalanced trees).

✨

Key Insights

1Insight 1: The perfect binary tree structure allows for a constant space solution by leveraging existing links to traverse the tree level by level.
2Insight 2: The solution can avoid explicit level-order traversal (BFS) using the `next` pointers to move horizontally across each level.
3Insight 3: Recognizing the pattern of connecting `left` to `right` child within the same node and `right` child to `left` child of the `next` node.

📚

Key Terms Explained

For Students

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The solution correctly handles this by returning null immediately.
Edge case 2: Single node tree. The while loop `while (leftmost.left)` will not execute, so the function returns the root, which is correct because the single node has no `next` node.
Edge case 3: A complete perfect binary tree. The algorithm correctly traverses and connects all nodes at each level from left to right.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the connection between the rightmost node of the left subtree and the leftmost node of the right subtree (i.e., `head.right.next = head.next.left`).

✗Common mistake 2: Not handling the edge case of an empty tree (root is null).

✗Common mistake 3: Incorrectly implementing the level-order traversal, leading to incorrect `next` pointer assignments.

✗Common mistake 4: Using extra space that isn't constant. Solutions using a queue or storing nodes in a list will be O(n) space.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This pattern of connecting nodes at the same level can be applied in graph algorithms where information needs to be propagated horizontally across nodes in a layered structure.

🏦Real-world scenario 2: Can be useful in implementing data structures with efficient horizontal traversal, such as distributed hash tables or peer-to-peer networks where nodes need to communicate with their neighbors on the same level.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The dummy node technique prevents edge cases. For cycle detection, Floyd's algorithm is the standard. Always handle empty list and single node cases.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Losing reference to nodes when modifying pointers. Use temporary variables to save references.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

Why use a dummy node in linked list problems?

A dummy node simplifies edge cases where the head might change (insertion/deletion at head) or when building a new list. Initialize dummy.next = head, work with dummy.next throughout, and return dummy.next at the end to handle all cases uniformly.

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Populating Next Right Pointers in Each Node"?

For this medium-level Linked List problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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