What is the time complexity of Convert Sorted List to Binary Search Tree?

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

What is the space complexity of Convert Sorted List to Binary Search Tree?

The space complexity is O(log(n)).

What data structures are used to solve Convert Sorted List to Binary Search Tree?

This problem uses Linked List, Divide and Conquer, Tree, Binary Search Tree, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Convert Sorted List to Binary Search Tree?

Insight 1: The inorder traversal of a BST yields a sorted sequence. Since we have a sorted linked list, we can simulate an inorder traversal during the tree construction process. Insight 2: To build a balanced BST, the middle element of the sorted list should be the root. Recursively constructing the left and right subtrees using the elements before and after the middle element will result in a balanced tree.

Which companies ask LeetCode 109. Convert Sorted List to Binary Search Tree in interviews?

This problem is frequently asked at Lyft, Meta, Uber, Zenefits.

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Convert Sorted List to Binary Search Tree - LeetCode 109 Solution

Q: How to solve Convert Sorted List to Binary Search Tree?

The provided solution uses a recursive approach combined with a 'global' pointer to the head of the linked list to mimic an inorder traversal while constructing the BST. First, the length of the list is computed. Then, a recursive helper function `build_tree` is defined, which takes the left and right boundaries of the sublist as input. In each recursive call, the middle element of the sublist is chosen as the root. The left and right subtrees are built recursively. A crucial optimization is maintaining a global `head_node` pointer that initially points to the head of the list. Before processing the right subtree, this `head_node` pointer is advanced to simulate the inorder traversal.

Convert Sorted List to Binary Search Tree - Complete Solution Guide

Convert Sorted List to Binary Search Tree is LeetCode problem 109, 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 head of a singly linked list where elements are sorted in ascending order , convert it to a height-balanced binary search tree . Example 1: Input: head = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST. Example 2: Input: head = [] Output: [] Constraints: The number of nodes in head is in the range [0, 2 * 10 4 ] . -10 5 <= Node.val <= 10 5

Detailed Explanation

The problem asks us to convert a sorted singly linked list into a height-balanced Binary Search Tree (BST). A height-balanced BST is a binary tree where the height difference between the left and right subtrees of every node is no more than one. The input is the head of a sorted linked list, and the output is the root of the height-balanced BST. The list is sorted in ascending order, which is crucial for building a BST.

Solution Approach

The provided solution uses a recursive approach combined with a 'global' pointer to the head of the linked list to mimic an inorder traversal while constructing the BST. First, the length of the list is computed. Then, a recursive helper function `build_tree` is defined, which takes the left and right boundaries of the sublist as input. In each recursive call, the middle element of the sublist is chosen as the root. The left and right subtrees are built recursively. A crucial optimization is maintaining a global `head_node` pointer that initially points to the head of the list. Before processing the right subtree, this `head_node` pointer is advanced to simulate the inorder traversal.

Step-by-Step Algorithm

Step 1: Calculate the length of the linked list. This is done by iterating through the list and incrementing a counter.
Step 2: Initialize a 'global' `head_node` pointer to the head of the linked list. This pointer will be used to traverse the list during the tree construction.
Step 3: Define a recursive helper function `build_tree(left, right)` that takes two integers `left` and `right` representing the start and end indices of the current sublist.
Step 4: In `build_tree`, check the base case: if `left > right`, return `None` (or `nullptr` in C++ or `NULL` in C or null in Java) as there's nothing to process.
Step 5: Calculate the middle index `mid = left + (right - left) // 2`. This avoids potential integer overflow.
Step 6: Recursively build the left subtree by calling `build_tree(left, mid - 1)`.
Step 7: Create a new `TreeNode` with the value of the current `head_node`. Assign the left subtree returned in Step 6 as the left child of the new node.
Step 8: Advance the `head_node` pointer to its next node in the linked list. This simulates the inorder traversal.
Step 9: Recursively build the right subtree by calling `build_tree(mid + 1, right)`. Assign the result as the right child of the current node.
Step 10: Return the current `TreeNode` (root).

Key Insights

Insight 1: The inorder traversal of a BST yields a sorted sequence. Since we have a sorted linked list, we can simulate an inorder traversal during the tree construction process.
Insight 2: To build a balanced BST, the middle element of the sorted list should be the root. Recursively constructing the left and right subtrees using the elements before and after the middle element will result in a balanced tree.
Insight 3: The key to solving this problem efficiently is avoiding repeatedly traversing the linked list to find the middle element in each recursive call. We can maintain a global pointer to the head of the linked list and advance it as we build the tree, simulating an inorder traversal.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(log(n))

Topics

This problem involves: Linked List, Divide and Conquer, Tree, Binary Search Tree, Binary Tree.

Companies

Asked at: Lyft, Meta, Uber, Zenefits.

Convert Sorted List to Binary Search Tree | LeetCode 109

Medium

Linked List Divide and Conquer Tree Binary Search Tree Binary Tree

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

Solution Code

Available in:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * 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 {

    private ListNode head_node;

    public TreeNode sortedListToBST(ListNode head) {
        ListNode current = head;
        int size = 0;
        while (current != null) {
            size += 1;
            current = current.next;
        }

        this.head_node = head;

        return build_tree(0, size - 1);
    }

    private TreeNode build_tree(int left, int right) {
        if (left > right) {
            return null;
        }

        int mid = left + (right - left) / 2;

        TreeNode left_child = build_tree(left, mid - 1);

        TreeNode root = new TreeNode(this.head_node.val);
        root.left = left_child;

        this.head_node = this.head_node.next;

        root.right = build_tree(mid + 1, right);

        return root;
    }
}

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Problem Info

Description

Problem 109: Convert Sorted List to Binary Search Tree - Detailed explanation and solution approaches.

Topics

Linked List Divide and Conquer Tree Binary Search Tree Binary Tree

Companies

LyftMetaUberZenefits

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:109

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

10 steps

1
Step 1: Calculate the length of the linked list. This is done by iterating through the list and incrementing a counter.
2
Step 2: Initialize a 'global' `head_node` pointer to the head of the linked list. This pointer will be used to traverse the list during the tree construction.
3
Step 3: Define a recursive helper function `build_tree(left, right)` that takes two integers `left` and `right` representing the start and end indices of the current sublist.
4
Step 4: In `build_tree`, check the base case: if `left > right`, return `None` (or `nullptr` in C++ or `NULL` in C or null in Java) as there's nothing to process.
5
Step 5: Calculate the middle index `mid = left + (right - left) // 2`. This avoids potential integer overflow.
6
Step 6: Recursively build the left subtree by calling `build_tree(left, mid - 1)`.
7
Step 7: Create a new `TreeNode` with the value of the current `head_node`. Assign the left subtree returned in Step 6 as the left child of the new node.
8
Step 8: Advance the `head_node` pointer to its next node in the linked list. This simulates the inorder traversal.
9
Step 9: Recursively build the right subtree by calling `build_tree(mid + 1, right)`. Assign the result as the right child of the current node.
10
Step 10: Return the current `TreeNode` (root).

💡

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(log(n))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The O(n) time complexity arises from processing each node in the linked list exactly once. The `build_tree` function divides the list conceptually in half but because we're only processing each node once as we move `head_node`, there is a linear dependency. The O(log n) space complexity is due to the recursion depth, which is proportional to the height of the balanced BST.

📖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

Convert to Array

Convert the linked list to a sorted array, and then use the array to construct the BST using a divide-and-conquer approach.

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

Trade-offs: This approach requires O(n) extra space to store the array. However, it might be easier to understand and implement for some. The original algorithm is more space-efficient. If memory is highly constrained, the solution using the linked list directly is better.

✨

Key Insights

1Insight 1: The inorder traversal of a BST yields a sorted sequence. Since we have a sorted linked list, we can simulate an inorder traversal during the tree construction process.
2Insight 2: To build a balanced BST, the middle element of the sorted list should be the root. Recursively constructing the left and right subtrees using the elements before and after the middle element will result in a balanced tree.
3Insight 3: The key to solving this problem efficiently is avoiding repeatedly traversing the linked list to find the middle element in each recursive call. We can maintain a global pointer to the head of the linked list and advance it as we build the tree, simulating an inorder traversal.

📚

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 list (head is None/null). The algorithm correctly handles this because the initial `size` will be 0, so `build_tree(0, -1)` will immediately return `None/null`. This is important because it provides the base case for empty subtrees as well.
Edge case 2: List with a single node. The algorithm correctly creates a tree with a single node, the list's head. `size` will be 1. build_tree(0,0) will create a tree node with `head_node.val` and left and right will be null as left will call build_tree(0,-1) and right will call build_tree(1,0), both returning null.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Repeatedly traversing the linked list to find the middle element. This would result in O(n log n) time complexity. The 'global' `head_node` pointer avoids this.

✗Common mistake 2: Not handling the base case of an empty list correctly. The algorithm needs to explicitly check for `head == None/null` at the beginning to avoid errors.

✗Common mistake 3: Integer overflow when calculating the middle index. Using `mid = left + (right - left) // 2` is safer than `mid = (left + right) // 2`.

✅

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: Creating an indexed search tree from a sorted data source. This is useful in databases where data is pre-sorted and needs to be accessed efficiently using a BST.

🏦Real-world scenario 2: Building a balanced index for quick lookups in sorted data structures. For example, converting a sorted log file into a structured format for efficient searching.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 18, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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.

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 "Convert Sorted List to Binary Search Tree"?

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 divide and conquer techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?