What is the time complexity of Reverse Odd Levels of Binary Tree?

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

What data structures are used to solve Reverse Odd Levels of Binary Tree?

This problem uses 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 Reverse Odd Levels of Binary Tree?

Insight 1: The problem is best solved using a recursive Depth-First Search (DFS) approach due to the hierarchical structure of the binary tree. Insight 2: We can traverse the tree, keeping track of the current level. When the level is odd, we swap the values of nodes that are symmetrical to each other with respect to the root.

Which companies ask LeetCode 2415. Reverse Odd Levels of Binary Tree in interviews?

This problem is frequently asked at J.P. Morgan, josh technology.

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Reverse Odd Levels of Binary Tree - LeetCode 2415 Solution

Reverse Odd Levels of Binary Tree - Complete Solution Guide

Reverse Odd Levels of Binary Tree is LeetCode problem 2415, 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 the root of a perfect binary tree, reverse the node values at each odd level of the tree. For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18] , then it should become [18,29,11,7,4,3,1,2] . Return the root of the reversed tree . A binary tree is perfect if all parent nodes have two children and all leaves are on the same level. The level of a node is the number of edges along the path between it and the root node. Example 1: Input: root = [2,3,5,8,13,21,34] Output: [2,5

Detailed Explanation

The problem asks us to reverse the node values at each odd level of a perfect binary tree. A perfect binary tree is one where all internal nodes have two children and all leaves are at the same level. The level of a node is defined as the number of edges between the node and the root. We need to reverse the values at level 1, 3, 5, and so on, while keeping the values at even levels unchanged. The root node is at level 0.

Solution Approach

The solution uses a recursive DFS approach to traverse the tree. The `dfs` function takes two nodes (one from the left subtree and one from the right subtree) and a boolean flag `is_odd_level` indicating whether the current level is odd. If `is_odd_level` is true, it swaps the values of the two nodes. Then, it recursively calls itself with the left child of the first node and the right child of the second node, and the right child of the first node and the left child of the second node, toggling the `is_odd_level` flag at each recursive call. This ensures that only nodes at odd levels have their values swapped.

Step-by-Step Algorithm

Step 1: Define a recursive helper function `dfs(node1, node2, is_odd_level)` to perform the tree traversal and value swapping.
Step 2: Base case: If either `node1` or `node2` is `None`, return because there's nothing to process.
Step 3: If `is_odd_level` is true, swap the values of `node1` and `node2`.
Step 4: Recursively call `dfs(node1.left, node2.right, !is_odd_level)` to process the left subtree of `node1` and the right subtree of `node2`.
Step 5: Recursively call `dfs(node1.right, node2.left, !is_odd_level)` to process the right subtree of `node1` and the left subtree of `node2`.
Step 6: In the main function, call `dfs(root.left, root.right, True)` to start the traversal with the left and right children of the root, setting `is_odd_level` to `True` initially since level 1 is odd.
Step 7: Return the modified `root` node.

Key Insights

Insight 1: The problem is best solved using a recursive Depth-First Search (DFS) approach due to the hierarchical structure of the binary tree.
Insight 2: We can traverse the tree, keeping track of the current level. When the level is odd, we swap the values of nodes that are symmetrical to each other with respect to the root.
Insight 3: Since the tree is perfect, for each node on an odd level, there exists a symmetrical node on the same level on the opposite side of the root. We can use this symmetry to swap the node values efficiently.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: J.P. Morgan, josh technology.

Reverse Odd Levels of Binary Tree | LeetCode 2415

Medium

Tree Depth-First Search Breadth-First Search Binary Tree

Time: O(n)Space: O(h)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode reverseOddLevels(TreeNode root) {
        
        function(root.left, root.right, true);
        return root;
    }

    public void function(TreeNode node1, TreeNode node2, boolean is_odd_level){
        if (node1 == null){
            return;
        }

        if (is_odd_level){
            int temp = node1.val;
            node1.val = node2.val;
            node2.val = temp;
        }

        function(node1.left, node2.right, !is_odd_level);
        function(node1.right, node2.left, !is_odd_level);
    }
}

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Problem Info

Description

Problem 2415: Reverse Odd Levels of Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

J.P. Morganjosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2415

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

7 steps

1
Step 1: Define a recursive helper function `dfs(node1, node2, is_odd_level)` to perform the tree traversal and value swapping.
2
Step 2: Base case: If either `node1` or `node2` is `None`, return because there's nothing to process.
3
Step 3: If `is_odd_level` is true, swap the values of `node1` and `node2`.
4
Step 4: Recursively call `dfs(node1.left, node2.right, !is_odd_level)` to process the left subtree of `node1` and the right subtree of `node2`.
5
Step 5: Recursively call `dfs(node1.right, node2.left, !is_odd_level)` to process the right subtree of `node1` and the left subtree of `node2`.
6
Step 6: In the main function, call `dfs(root.left, root.right, True)` to start the traversal with the left and right children of the root, setting `is_odd_level` to `True` initially since level 1 is odd.
7
Step 7: Return the modified `root` node.

💡

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(h)

📝

Why?

The time complexity is O(n) because each node is visited once during the DFS traversal. The space complexity is O(h) due to the maximum depth of the recursive call stack, which corresponds to the height of the tree. The stack frames store information like the current node, the other symmetrical node, and whether it is an odd level.

📖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)

An alternative approach is to use BFS (level-order traversal). We can iterate through the tree level by level. When we reach an odd level, we can store the node values in a list, reverse the list, and then update the node values with the reversed list.

Time: O(n), where n is the number of nodes in the tree.Space: O(w), where w is the maximum width of the tree. In a perfect binary tree, the maximum width is n/2 (at the last level), so it can also be considered O(n).

Trade-offs: While the time complexity is the same, BFS may use more space than DFS, especially for very wide trees. DFS is generally preferred because its space complexity is based on the height of the tree, which is typically logarithmic for balanced trees, while BFS space complexity depends on the tree's width.

✨

Key Insights

1Insight 1: The problem is best solved using a recursive Depth-First Search (DFS) approach due to the hierarchical structure of the binary tree.
2Insight 2: We can traverse the tree, keeping track of the current level. When the level is odd, we swap the values of nodes that are symmetrical to each other with respect to the root.
3Insight 3: Since the tree is perfect, for each node on an odd level, there exists a symmetrical node on the same level on the opposite side of the root. We can use this symmetry to swap the node values efficiently.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is None). The solution handles this gracefully because the initial check `if root:` prevents any operations if the tree is empty.
Edge case 2: Tree with only the root node. The DFS function is never called as it starts with the root's children, so the root remains unchanged, as desired.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the odd level condition correctly. Forgetting to toggle the `is_odd_level` flag in the recursive calls leads to incorrect reversals.

✗Common mistake 2: Incorrectly swapping values, which can lead to incorrect values for nodes in later levels. The symmetrical swap using two node references is crucial.

✗Common mistake 3: Failing to recognize and take advantage of the perfect tree structure, which simplifies the symmetrical node identification.

✅

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: Image processing where you might need to mirror or invert certain levels of a hierarchical image representation (e.g., an octree).

🏦Real-world scenario 2: In database indexing using tree-like structures, you might need to reorganize or rebalance specific levels based on certain criteria, conceptually similar to reversing levels.

💡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

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

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

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: August 15, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

Frequently Asked Questions

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 "Reverse Odd Levels of Binary Tree"?

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

🤖 Need help with this problem?