What is the time complexity of Sum of Left Leaves?

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

What data structures are used to solve Sum of Left Leaves?

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 Sum of Left Leaves?

Insight 1: The definition of a 'left leaf' is crucial. We need to identify nodes that are leaves AND are the left children of another node. Insight 2: A recursive approach is well-suited for traversing the binary tree and checking each node.

Which companies ask LeetCode 404. Sum of Left Leaves in interviews?

This problem is frequently asked at Grammarly.

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Sum of Left Leaves - LeetCode 404 Solution

Q: How to solve Sum of Left Leaves?

The provided solution uses a recursive depth-first search (DFS) approach to traverse the binary tree. For each node, it checks if its left child is a leaf node. If it is, the value of the left child is added to the sum. The function then recursively calls itself on the left and right subtrees, accumulating the sum of left leaves from both subtrees. The final sum is returned.

Sum of Left Leaves - Complete Solution Guide

Sum of Left Leaves is LeetCode problem 404, 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 sum of all left leaves. A leaf is a node with no children. A left leaf is a leaf that is the left child of another node. Example 1: Input: root = [3,9,20,null,null,15,7] Output: 24 Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively. Example 2: Input: root = [1] Output: 0 Constraints: The number of nodes in the tree is in the range [1, 1000] . -1000 <= Node.val <= 1000

Detailed Explanation

The problem asks us to find the sum of all left leaves in a given binary tree. A left leaf is defined as a node that is both a leaf (has no children) and is the left child of its parent. The input is the root of the binary tree, and the output is an integer representing the sum of the values of all left leaves in the tree.

Solution Approach

The provided solution uses a recursive depth-first search (DFS) approach to traverse the binary tree. For each node, it checks if its left child is a leaf node. If it is, the value of the left child is added to the sum. The function then recursively calls itself on the left and right subtrees, accumulating the sum of left leaves from both subtrees. The final sum is returned.

Step-by-Step Algorithm

Step 1: Check if the current node (root) is null. If it is, return 0 (base case).
Step 2: Initialize a variable `sum` to 0. This variable will store the sum of left leaves.
Step 3: Check if the current node has a left child. If it does, proceed to the next step; otherwise, skip to step 5.
Step 4: If the left child is a leaf node (i.e., it has no left and no right children), add its value to `sum`. This is the key step to identify and count left leaves.
Step 5: Recursively call the `sumOfLeftLeaves` function on the left subtree and add the returned value to `sum`.
Step 6: Recursively call the `sumOfLeftLeaves` function on the right subtree and add the returned value to `sum`.
Step 7: Return the final `sum`.

Key Insights

Insight 1: The definition of a 'left leaf' is crucial. We need to identify nodes that are leaves AND are the left children of another node.
Insight 2: A recursive approach is well-suited for traversing the binary tree and checking each node.
Insight 3: Base case for recursion: If the root is null, the sum of left leaves is 0.

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: Grammarly.

Sum of Left Leaves | LeetCode 404

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 sumOfLeftLeaves(TreeNode root) {
        if(root == null) return 0;
        int sum = 0;
        if(root.left != null && root.left.left == null && root.left.right == null){
            sum += root.left.val;
        }
        sum += sumOfLeftLeaves(root.left);
        sum += sumOfLeftLeaves(root.right);
        return sum;
    }
}

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Problem Info

Description

Problem 404: Sum of Left Leaves - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

Grammarly

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:404

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: Check if the current node (root) is null. If it is, return 0 (base case).
2
Step 2: Initialize a variable `sum` to 0. This variable will store the sum of left leaves.
3
Step 3: Check if the current node has a left child. If it does, proceed to the next step; otherwise, skip to step 5.
4
Step 4: If the left child is a leaf node (i.e., it has no left and no right children), add its value to `sum`. This is the key step to identify and count left leaves.
5
Step 5: Recursively call the `sumOfLeftLeaves` function on the left subtree and add the returned value to `sum`.
6
Step 6: Recursively call the `sumOfLeftLeaves` function on the right subtree and add the returned value to `sum`.
7
Step 7: Return the final `sum`.

💡

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 we must potentially visit every node in the tree. The space complexity stems from the recursion depth. In the worst case of a completely unbalanced tree (e.g., a linked list), the depth will be O(n). In the best case of a balanced tree, the depth is O(log n).

📖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 Approach using Stack (DFS)

An iterative solution can be implemented using a stack to mimic the DFS traversal. Instead of recursive calls, the stack keeps track of the nodes to visit.

Time: O(n)Space: O(h) where h is the height of the tree. In the worst case, h can be n for a skewed tree.

Trade-offs: Iterative solutions avoid recursion overhead and can sometimes be more efficient, but they might be more complex to implement and understand compared to recursive solutions for tree traversals.

Iterative Approach using Queue (BFS)

An iterative solution can be implemented using a queue to perform Breadth-First Search traversal. Although this approach is valid, BFS is less intuitive for solving this particular problem than DFS. You would need to keep track of the parent node to check for left leaves.

Time: O(n)Space: O(w) where w is the maximum width of the tree. In the worst case (balanced tree), w can be n/2.

Trade-offs: BFS offers a different traversal order but is less natural than DFS for this problem. The space complexity can potentially be higher than DFS for balanced trees.

✨

Key Insights

1Insight 1: The definition of a 'left leaf' is crucial. We need to identify nodes that are leaves AND are the left children of another node.
2Insight 2: A recursive approach is well-suited for traversing the binary tree and checking each node.
3Insight 3: Base case for recursion: If the root is null, the sum of left leaves is 0.

📚

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 case by returning 0.
Edge case 2: A tree with only a root node. The solution correctly returns 0 because there can be no left leaves.
Edge case 3: A tree where the root has only a right child. The solution correctly returns 0 as there's no left subtree to examine.
Edge case 4: A tree where a node has a left child which is a leaf, and a right child which is not a leaf. The code accurately identifies and adds only the left leaf's value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if a node is a leaf node before adding its value. The problem specifically requires summing the values of *left leaves*, not just any left node.

✗Common mistake 2: Not handling the base case (empty tree) correctly, which can lead to null pointer exceptions or incorrect results.

✗Common mistake 3: Incorrectly identifying left leaves by only checking if a node is the left child of its parent, without also verifying that it has no children of its own.

✅

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 structures as trees. Summing specific values (e.g., project costs) associated with certain 'leaf' departments on the left side of the hierarchy could be analogous to this problem.

🏦Real-world scenario 2: In compiler design, abstract syntax trees (ASTs) are used. Identifying and summing attributes of certain leaf nodes on the left-hand side of expressions can be similar to this problem. For example, summing the integer values of left-most terminal nodes in an arithmetic expression.

💡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: July 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

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 "Sum of Left Leaves"?

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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