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

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

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

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 II?

Insight 1: The problem can be solved using a level-order traversal, connecting nodes as we traverse each level. Insight 2: Since the follow-up requires constant extra space, we need an approach that avoids using a queue or list for level-order traversal.

Which companies ask LeetCode 117. Populating Next Right Pointers in Each Node II in interviews?

This problem is frequently asked at Citadel, Snowflake, Walmart Labs.

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Populating Next Right Pointers in Each Node II - LeetCode 117 Solution

Populating Next Right Pointers in Each Node II - Complete Solution Guide

Populating Next Right Pointers in Each Node II is LeetCode problem 117, 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

Given a binary tree 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,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 seria

Detailed Explanation

The problem asks us to connect all nodes at the same level in a given binary tree using the 'next' pointer. Each node in the tree has a 'val', 'left', 'right', and 'next' field. Initially, all 'next' pointers are NULL. The goal is to populate the 'next' pointers so that each node points to the node to its right on the same level. If a node is the rightmost node on its level, its 'next' pointer should remain NULL. The input is the root node of the binary tree, and the output is the root of the modified tree with the 'next' pointers correctly populated.

Solution Approach

The provided solution uses a level-order traversal approach but cleverly avoids using extra space. It uses the existing 'next' pointers of the previously processed level to link nodes on the current level. The outer `while` loop iterates through each level of the tree. Inside this loop, a dummy node is used to keep track of the beginning of the new level that we are creating the 'next' links for. The inner `while` loop iterates through the nodes of the current level using the 'next' pointers established in the previous level, and creates the next level's 'next' links to point left to right.

Step-by-Step Algorithm

Step 1: Start with the root node as the head of the first level.
Step 2: While the 'head' is not NULL (i.e., there are more levels to process), create a dummy node and a tail pointer.
Step 3: Initialize a 'curr' pointer to the 'head'. This pointer traverses nodes in the current level using the existing 'next' pointers.
Step 4: While the 'curr' pointer is not NULL, check if the current node has a left child. If it does, append it to the 'tail' pointer and advance the 'tail'.
Step 5: Check if the current node has a right child. If it does, append it to the 'tail' pointer and advance the 'tail'.
Step 6: Move to the next node in the current level using `curr = curr.next`.
Step 7: After processing all nodes on the current level, update the 'head' to be the first node of the next level (dummy.next).
Step 8: Repeat steps 2-7 until 'head' becomes NULL.

Key Insights

Insight 1: The problem can be solved using a level-order traversal, connecting nodes as we traverse each level.
Insight 2: Since the follow-up requires constant extra space, we need an approach that avoids using a queue or list for level-order traversal.
Insight 3: Utilize the existing 'next' pointers of the previous level to traverse the current level.

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, Snowflake, Walmart Labs.

Populating Next Right Pointers in Each Node II | LeetCode 117

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 root;
        }

        Node head = root;

        while (head != null) {
            Node dummy = new Node(0);
            Node tail = dummy;
            Node curr = head;

            while (curr != null) {
                if (curr.left != null) {
                    tail.next = curr.left;
                    tail = tail.next;
                }
                if (curr.right != null) {
                    tail.next = curr.right;
                    tail = tail.next;
                }
                curr = curr.next;
            }

            head = dummy.next;
        }

        return root;
    }
}

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Problem Info

Description

Problem 117: Populating Next Right Pointers in Each Node II - Detailed explanation and solution approaches.

Topics

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

Companies

CitadelSnowflakeWalmart Labs

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:117

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

8 steps

1
Step 1: Start with the root node as the head of the first level.
2
Step 2: While the 'head' is not NULL (i.e., there are more levels to process), create a dummy node and a tail pointer.
3
Step 3: Initialize a 'curr' pointer to the 'head'. This pointer traverses nodes in the current level using the existing 'next' pointers.
4
Step 4: While the 'curr' pointer is not NULL, check if the current node has a left child. If it does, append it to the 'tail' pointer and advance the 'tail'.
5
Step 5: Check if the current node has a right child. If it does, append it to the 'tail' pointer and advance the 'tail'.
6
Step 6: Move to the next node in the current level using `curr = curr.next`.
7
Step 7: After processing all nodes on the current level, update the 'head' to be the first node of the next level (dummy.next).
8
Step 8: Repeat steps 2-7 until 'head' becomes NULL.

💡

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 a constant number of times. The outer loop visits each level, and the inner loop iterates through the nodes on that level, processing each node in the tree exactly once. Space complexity is O(1) because the algorithm only uses a fixed number of pointers, regardless of the input size.

📖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

A standard BFS approach using a queue. Iterate level by level and connect adjacent nodes.

Time: O(n)Space: O(w), where w is the maximum width of the tree (worst case O(n) for a skewed tree).

Trade-offs: Simpler to understand initially, but violates the constant space constraint. Useful if space is not a concern.

Recursive Approach

A recursive approach can be used to traverse the tree, but maintaining the 'next' pointers within the recursion can be complex and doesn't easily lend itself to constant space.

Time: O(n)Space: O(h), where h is the height of the tree (O(log n) for a balanced tree, O(n) for a skewed tree). While the prompt mentioned that the recursive stack space does not count, it is still important to keep in mind in a real world environment.

Trade-offs: The recursive approach might be more readable for some, but it doesn't meet the constant space requirement and it is harder to get right without using extra space.

✨

Key Insights

1Insight 1: The problem can be solved using a level-order traversal, connecting nodes as we traverse each level.
2Insight 2: Since the follow-up requires constant extra space, we need an approach that avoids using a queue or list for level-order traversal.
3Insight 3: Utilize the existing 'next' pointers of the previous level to traverse the current level.

📚

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 handles this by immediately returning the root.
Edge case 2: Tree with only one node. The while loop condition `head` will be NULL immediately, so root is returned.
Edge case 3: Skewed tree (only left or right children). The algorithm correctly connects nodes level by level, even if some levels have only one node.
Edge case 4: A complete binary tree. The algorithm will correctly connect all the nodes at the same level.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case when a node has only a left or only a right child. The solution must correctly handle both cases.

✗Common mistake 2: Using extra space (e.g., a queue) when the problem explicitly asks for a constant space solution. A queue is the obvious solution, but cannot be used.

✗Common mistake 3: Incorrectly updating the 'head' pointer after each level, leading to an infinite loop or incorrect connections.

✗Common mistake 4: Forgetting to initialize the `next` pointer of the dummy node to null.

✅

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: Representing organizational charts where employees at the same level need to be easily traversed. The 'next' pointers would allow for quick iteration across levels.

🏦Real-world scenario 2: In parallel processing, distributing tasks to nodes at the same level in a tree structure. The 'next' pointers could facilitate efficient communication between sibling nodes.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 26, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

How does the fast/slow pointer technique work?

Fast pointer moves 2 steps while slow moves 1 step. When fast reaches the end, slow is at middle. If there's a cycle, they will meet. This technique is used for: finding middle node, detecting cycles, finding cycle start point, and checking palindromes.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

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.

🤖 Need help with this problem?