What is the time complexity of Binary Tree Postorder Traversal?

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

What data structures are used to solve Binary Tree Postorder Traversal?

This problem uses Stack, Tree, Depth-First Search, Binary Tree. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Binary Tree Postorder Traversal?

Insight 1: Postorder traversal inherently requires visiting nodes in a specific order: left, right, then root. This suggests a recursive approach or an iterative approach using a stack to mimic the recursive call stack. Insight 2: Using a stack is crucial for the iterative approach. The stack keeps track of nodes to visit, ensuring the correct order of traversal.

Which companies ask LeetCode 145. Binary Tree Postorder Traversal in interviews?

This problem is frequently asked at Google.

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Binary Tree Postorder Traversal - LeetCode 145 Solution

Q: How to solve Binary Tree Postorder Traversal?

The provided solutions use iterative approaches employing a stack to perform the postorder traversal. The Python solution uses a simple stack, pushing nodes and then reversing the result to obtain the postorder sequence. The Java and C++ solutions utilize a more sophisticated iterative approach that avoids the need for reversing the result. This optimized iterative approach cleverly uses a `lastVisited` variable to track the last visited node, ensuring the correct order.

Binary Tree Postorder Traversal - Complete Solution Guide

Binary Tree Postorder Traversal is LeetCode problem 145, a Easy 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 binary tree, return the postorder traversal of its nodes' values . Example 1: Input: root = [1,null,2,3] Output: [3,2,1] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [4,6,7,5,2,9,8,3,1] Explanation: Example 3: Input: root = [] Output: [] Example 4: Input: root = [1] Output: [1] Constraints: The number of the nodes in the tree is in the range [0, 100] . -100 <= Node.val <= 100 Follow up: Recursive solution is trivial, could you do it iterati

Detailed Explanation

The problem asks for a postorder traversal of a binary tree. Postorder traversal means visiting the left subtree, then the right subtree, and finally the root node. The input is the root node of a binary tree, and the output is a list containing the values of the nodes in postorder sequence. The tree can be empty or contain any number of nodes within the specified constraints. Constraints limit the number of nodes and the range of values within each node.

Solution Approach

The provided solutions use iterative approaches employing a stack to perform the postorder traversal. The Python solution uses a simple stack, pushing nodes and then reversing the result to obtain the postorder sequence. The Java and C++ solutions utilize a more sophisticated iterative approach that avoids the need for reversing the result. This optimized iterative approach cleverly uses a `lastVisited` variable to track the last visited node, ensuring the correct order.

Step-by-Step Algorithm

Step 1: Initialize an empty list/array `result` to store the postorder traversal sequence and a stack `stack`.
Step 2: Handle the base case: If the root is `null` (empty tree), return the empty `result`.
Step 3: (Python): Push the root node onto the stack. While the stack is not empty, pop a node, append its value to `result`, and push its left and right children (if they exist) onto the stack. Finally, reverse `result`.
Step 4: (Java/C++): Push the root node onto the stack. While the stack is not empty or the current node `curr` is not null: If `curr` is not null, push it onto the stack and move to its left child. Otherwise, examine the top node of the stack: If it has a right child that hasn't been visited, move to that right child. Otherwise, add the top node's value to `result`, and pop it. This mimics the recursive call stack precisely.
Step 5: Return the `result` list/array.

Key Insights

Insight 1: Postorder traversal inherently requires visiting nodes in a specific order: left, right, then root. This suggests a recursive approach or an iterative approach using a stack to mimic the recursive call stack.
Insight 2: Using a stack is crucial for the iterative approach. The stack keeps track of nodes to visit, ensuring the correct order of traversal.
Insight 3: Handling the empty tree case is important. A robust solution should gracefully handle an empty input without causing errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Google.

Binary Tree Postorder Traversal | LeetCode 145

Easy

Stack Tree Depth-First Search Binary Tree

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

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 List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }

        Stack<TreeNode> stack = new Stack<>();
        TreeNode lastVisited = null;
        TreeNode curr = root;

        while (!stack.isEmpty() || curr != null) {
            if (curr != null) {
                stack.push(curr);
                curr = curr.left;
            } else {
                TreeNode peekNode = stack.peek();
                if (peekNode.right != null && lastVisited != peekNode.right) {
                    curr = peekNode.right;
                } else {
                    result.add(peekNode.val);
                    lastVisited = stack.pop();
                }
            }
        }
        return result;
    }
}

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Problem Info

Description

Problem 145: Binary Tree Postorder Traversal - Detailed explanation and solution approaches.

Topics

Stack Tree Depth-First Search Binary Tree

Companies

Google

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:145

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

5 steps

1
Step 1: Initialize an empty list/array `result` to store the postorder traversal sequence and a stack `stack`.
2
Step 2: Handle the base case: If the root is `null` (empty tree), return the empty `result`.
3
Step 3: (Python): Push the root node onto the stack. While the stack is not empty, pop a node, append its value to `result`, and push its left and right children (if they exist) onto the stack. Finally, reverse `result`.
4
Step 4: (Java/C++): Push the root node onto the stack. While the stack is not empty or the current node `curr` is not null: If `curr` is not null, push it onto the stack and move to its left child. Otherwise, examine the top node of the stack: If it has a right child that hasn't been visited, move to that right child. Otherwise, add the top node's value to `result`, and pop it. This mimics the recursive call stack precisely.
5
Step 5: Return the `result` list/array.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because each node in the tree is visited exactly once. The space complexity is O(n) in the worst case (a skewed tree) because the stack can hold up to n nodes (height of the tree can be n in the worst case).

📖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

Recursive Approach

A recursive solution naturally mirrors the definition of postorder traversal. It would recursively traverse the left subtree, then the right subtree, and finally process the root node.

Time: O(n)Space: O(h) where h is the height of the tree (can be O(n) for a skewed tree)

Trade-offs: The recursive approach is more concise but might lead to stack overflow errors for very deep trees. The iterative approach is generally preferred for its guaranteed space efficiency.

✨

Key Insights

1Insight 1: Postorder traversal inherently requires visiting nodes in a specific order: left, right, then root. This suggests a recursive approach or an iterative approach using a stack to mimic the recursive call stack.
2Insight 2: Using a stack is crucial for the iterative approach. The stack keeps track of nodes to visit, ensuring the correct order of traversal.
3Insight 3: Handling the empty tree case is important. A robust solution should gracefully handle an empty input without causing errors.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree – The solutions correctly handle the case where the root is null, returning an empty list.
Edge case 2: Single-node tree – The solutions correctly handle a tree with only one node, returning a list containing that node's value.
Edge case 3: Skewed tree – The solutions perform correctly on trees that are heavily skewed to one side (either left or right), demonstrating efficient handling of depth.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect order of visiting nodes. Beginners often mix up the order of visiting left, right, and root in their traversal logic.

✗Common mistake 2: Not handling the base case (empty tree) correctly. Forgetting to check for a null root can lead to null pointer exceptions.

✅

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: Evaluating expressions represented as binary trees (e.g., evaluating arithmetic expressions). Postorder traversal is crucial because the operations are usually performed after evaluating the operands.

🏦Real-world scenario 2: File system traversal. Imagine a file system represented as a tree; postorder traversal would allow you to delete all files within a directory before deleting the directory itself.

💡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

Stacks are perfect for problems involving matching (parentheses) or "next greater element" patterns. If you need to look back at recent elements, think stack.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Popping from an empty stack. Always check if the stack is non-empty before popping.

👨‍💻 From Experience

This is one of the first problems I solved when learning Stack. It's a great foundation for more complex problems.

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: October 24, 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 a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Binary Tree Postorder Traversal"?

For this easy-level Stack 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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