What is the time complexity of Maximum Depth of Binary Tree?

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

What data structures are used to solve Maximum Depth of Binary Tree?

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

What are the key insights for solving Maximum Depth of Binary Tree?

**Recursive Definition for Depth:** The most critical insight is realizing that the depth of any non-empty node can be defined recursively: `Depth(node) = 1 + max(Depth(node.left), Depth(node.right))`. This precise mathematical relationship is the backbone of the efficient recursive solution, directly translating into code. **Explicit Base Case for Null Nodes:** Handling the `if not root: return 0` case is more than just an edge case; it's the fundamental termination condition for the recursion. It correctly assigns a depth of zero to an empty branch, preventing infinite recursion and providing the 'ground truth' for depth calculations when a leaf node's children are queried.

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Maximum Depth of Binary Tree - LeetCode 104 Solution

Q: How to solve Maximum Depth of Binary Tree?

The provided solution leverages a recursive Depth-First Search (DFS) approach, which is wonderfully intuitive for problems involving tree depths. The fundamental idea is that the maximum depth of any tree (or subtree) is determined by its deepest child branch. For a given `root` node, its maximum depth is `1` (for itself) plus the maximum depth found in either its left or right subtree. The algorithm works like this: First, it establishes a crucial base case: if the `root` is `None` (meaning an empty tree or we've traversed past a leaf node), its depth is `0`. This handles both an empty input tree and the termination condition for recursive calls. Otherwise, if there's a node, the function recursively calls itself on the `root.left` and `root.right` children to find their respective maximum depths. Once those values (`left_depth` and `right_depth`) return, the current node simply takes the maximum of these two depths and adds `1` (to account for itself) before returning. This elegant recursive decomposition perfectly mirrors the definition of maximum depth.

Q: Which companies ask LeetCode 104. Maximum Depth of Binary Tree in interviews?

This problem is frequently asked at Accenture, Adobe, Apple, Arista Networks, Bloomberg and 10 more companies.

Maximum Depth of Binary Tree - Complete Solution Guide

Maximum Depth of Binary Tree is LeetCode problem 104, 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 its maximum depth . A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Example 1: Input: root = [3,9,20,null,null,15,7] Output: 3 Example 2: Input: root = [1,null,2] Output: 2 Constraints: The number of nodes in the tree is in the range [0, 10 4 ] . -100 <= Node.val <= 100

Detailed Explanation

You're tasked with finding the 'maximum depth' of a given binary tree. It might sound straightforward, but let's be precise about what "maximum depth" means here. It's defined as the number of nodes you'd traverse along the *longest path* starting from the `root` node and ending at the *farthest leaf* node. Crucially, we're counting *nodes*, not edges. So, an empty tree has a depth of 0, and a tree with just a single root node has a depth of 1. This distinction is vital for correctly interpreting the problem statement.

Solution Approach

The provided solution leverages a recursive Depth-First Search (DFS) approach, which is wonderfully intuitive for problems involving tree depths. The fundamental idea is that the maximum depth of any tree (or subtree) is determined by its deepest child branch. For a given `root` node, its maximum depth is `1` (for itself) plus the maximum depth found in either its left or right subtree. The algorithm works like this: First, it establishes a crucial base case: if the `root` is `None` (meaning an empty tree or we've traversed past a leaf node), its depth is `0`. This handles both an empty input tree and the termination condition for recursive calls. Otherwise, if there's a node, the function recursively calls itself on the `root.left` and `root.right` children to find their respective maximum depths. Once those values (`left_depth` and `right_depth`) return, the current node simply takes the maximum of these two depths and adds `1` (to account for itself) before returning. This elegant recursive decomposition perfectly mirrors the definition of maximum depth.

Step-by-Step Algorithm

Step 1: Check if the root node is null. If it is, return 0 (base case: empty tree).
Step 2: Recursively call `maxDepth` on the left subtree to get its depth (`leftDepth`).
Step 3: Recursively call `maxDepth` on the right subtree to get its depth (`rightDepth`).
Step 4: Return the maximum of `leftDepth` and `rightDepth`, plus 1 (to account for the current node).

Key Insights

**Recursive Definition for Depth:** The most critical insight is realizing that the depth of any non-empty node can be defined recursively: `Depth(node) = 1 + max(Depth(node.left), Depth(node.right))`. This precise mathematical relationship is the backbone of the efficient recursive solution, directly translating into code.
**Explicit Base Case for Null Nodes:** Handling the `if not root: return 0` case is more than just an edge case; it's the fundamental termination condition for the recursion. It correctly assigns a depth of zero to an empty branch, preventing infinite recursion and providing the 'ground truth' for depth calculations when a leaf node's children are queried.
**Modeling Tree Structure with Recursion:** This problem beautifully showcases how recursion naturally aligns with the hierarchical structure of a tree. Each recursive call essentially processes a subtree, asking the same question ("What's your maximum depth?") to its children, allowing the results to bubble up and consolidate at the root without complex iterative state management.

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: Accenture, Adobe, Apple, Arista Networks, Bloomberg, Infosys, LinkedIn, Meta, Microsoft, Qualcomm, SAP, Spotify, Uber, Yahoo, Yandex.

Maximum Depth of Binary Tree | LeetCode 104

Easy

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 int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        } else {
            int leftDepth = maxDepth(root.left);
            int rightDepth = maxDepth(root.right);
            return Math.max(leftDepth, rightDepth) + 1;
        }
    }
}

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Problem Info

Description

Problem 104: Maximum Depth of Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

AccentureAdobeAppleArista NetworksBloombergInfosysLinkedInMeta

Stats

Difficulty:Easy

Languages:4

Companies:15

Problem #:104

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

4 steps

1
Step 1: Check if the root node is null. If it is, return 0 (base case: empty tree).
2
Step 2: Recursively call `maxDepth` on the left subtree to get its depth (`leftDepth`).
3
Step 3: Recursively call `maxDepth` on the right subtree to get its depth (`rightDepth`).
4
Step 4: Return the maximum of `leftDepth` and `rightDepth`, plus 1 (to account for the current 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?

Time complexity is O(n) because the algorithm visits each node in the tree exactly once. The space complexity is O(h) where h is the height of the tree, due to the recursive call stack. In the worst-case scenario (a skewed tree), h can be equal to n (number of nodes), resulting in O(n) space complexity. In a balanced tree, h is log(n), giving O(log n) space.

📖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

Iterative BFS

Use a queue to perform a breadth-first search. Level-order traversal counts levels until all nodes at a level are processed.

Time: O(n)Space: O(w), where w is the maximum width of the tree (can be O(n) in worst case)

Trade-offs: Iterative BFS might be slightly less efficient in space for very deep, narrow trees, but generally comparable to the recursive DFS approach. Iterative approach avoids potential stack overflow issues for extremely deep trees.

✨

Key Insights

1**Recursive Definition for Depth:** The most critical insight is realizing that the depth of any non-empty node can be defined recursively: `Depth(node) = 1 + max(Depth(node.left), Depth(node.right))`. This precise mathematical relationship is the backbone of the efficient recursive solution, directly translating into code.
2**Explicit Base Case for Null Nodes:** Handling the `if not root: return 0` case is more than just an edge case; it's the fundamental termination condition for the recursion. It correctly assigns a depth of zero to an empty branch, preventing infinite recursion and providing the 'ground truth' for depth calculations when a leaf node's children are queried.
3**Modeling Tree Structure with Recursion:** This problem beautifully showcases how recursion naturally aligns with the hierarchical structure of a tree. Each recursive call essentially processes a subtree, asking the same question ("What's your maximum depth?") to its children, allowing the results to bubble up and consolidate at the root without complex iterative state management.

📚

Key Terms Explained

For Students

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 0.
Edge case 2: Tree with only one node (root). The solution correctly returns 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the base case (empty tree). This leads to infinite recursion and a stack overflow error.

✗Common mistake 2: Incorrectly calculating the depth. Forgetting to add 1 in the recursive step to account for the current node.

✅

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: Analyzing the hierarchical structure of data, such as organizational charts, file systems, or decision trees. The maximum depth represents the maximum number of levels in the hierarchy.

🏦Real-world scenario 2: Determining the maximum number of steps required in a process that's represented as a tree structure (e.g., network routing, game tree search).

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

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

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 13, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

❓

Frequently Asked Questions

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Maximum Depth of Binary Tree"?

For this easy-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.

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