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Uber VO Real Questions｜Kth Largest Element in BST + Reverse In-Order Traversal Complete Analysis

2026-03-16

Uber VO Real Questions｜Kth Largest Element in BST + Reverse In-Order Traversal Complete Analysis

Introduction: Uber's VO (Virtual Onsite) interviews frequently test deep understanding of tree data structures. "Find the kth largest element in a BST" is a classic problem that evaluates candidates' understanding of BST properties, recursive traversal mastery, and ability to clarify problems and optimize solutions.

This article provides an in-depth review of this real question, including key clarifications during the interview, algorithm design, complexity analysis, and optimization strategies for handling multiple k values.

📌 Problem Description

Problem Statement

Given a Binary Search Tree (BST) and a positive integer k, find the kth largest element in the BST.

Example:

        10
       /  \
      4    20
     / \     \
    2   5     40

If k = 3, output should be 15
If k = 5, output should be 4

Descending Order: 40 > 20 > 15 > 10 > 5 > 4 > 2

🎯 Key Clarifications During Interview

Clarification 1: Tree Size Assumption

Question: Can we assume the binary tree contains at least k elements?

Interviewer's Answer: Yes, we can assume the tree contains at least k elements.

Significance: This simplifies the problem; we don't need to handle cases where k exceeds the tree size.

Clarification 2: Error Handling

Question: What should be returned if the tree size is less than k?

Candidate's Suggestion: Returning -1 could be confusing since -1 might be a valid value in the tree. Suggest throwing an error instead.

Interviewer's Agreement: This is a good observation, demonstrating careful thinking.

Clarification 3: Data Structure Definition

Question: Do we need to define a TreeNode class?

Interviewer's Answer: Yes, defining the TreeNode class is necessary to construct and manipulate the BST.

💡 Algorithm Design

Core Idea: Reverse In-Order Traversal

Leverage BST property: In-order traversal produces ascending sequence. Therefore, reverse in-order traversal produces descending sequence.

Reverse In-Order Traversal Order: Right Subtree → Root → Left Subtree

Detailed Steps

Using the example tree to find the 3rd largest element:

1. Start at root node 10, move right to 20
2. Continue moving right to 40
3. 40 has no right child, visit 40 (count = 1)
4. Back to 20, visit 20 (count = 2)
5. Move to left child of 20, which is 15, visit 15 (count = 3) ✓ Found!

Python Complete Code

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class KthLargestFinder:
    def __init__(self):
        self.count = 0
        self.result = None
    
    def kthLargest(self, root, k):
        """
        Find the kth largest element in BST
        
        Time Complexity: O(n) (worst case)
        Space Complexity: O(h) (recursion stack depth)
        """
        self.count = 0
        self.result = None
        self._reverse_inorder(root, k)
        
        if self.result is None:
            raise ValueError(f"No kth largest element for k={k}")
        
        return self.result
    
    def _reverse_inorder(self, node, k):
        """Reverse in-order traversal"""
        if node is None:
            return
        
        # Traverse right subtree first (larger values)
        self._reverse_inorder(node.right, k)
        
        # Visit current node
        self.count += 1
        if self.count == k:
            self.result = node.val
            return
        
        # Traverse left subtree (smaller values)
        self._reverse_inorder(node.left, k)

# Test Cases
if __name__ == "__main__":
    # Build example tree
    root = TreeNode(10)
    root.left = TreeNode(4)
    root.right = TreeNode(20)
    root.left.left = TreeNode(2)
    root.left.right = TreeNode(5)
    root.right.right = TreeNode(40)
    
    finder = KthLargestFinder()
    
    print(f"3rd largest element: {finder.kthLargest(root, 3)}")  # Output: 15
    print(f"5th largest element: {finder.kthLargest(root, 5)}")  # Output: 4

📊 Complexity Analysis

Metric Complexity Notes
Time Complexity O(n) Worst case requires traversing all nodes
Space Complexity O(h) Recursion stack depth equals tree height

Optimization: If the tree is balanced, h = O(log n), so space complexity becomes O(log n).

🔥 Extension: Multiple k Values

Problem

How to find elements for multiple k values in a single traversal?

Solution

def kthLargestMultiple(root, k_values):
    """
    Find multiple kth largest elements in one traversal
    
    Time Complexity: O(n)
    Space Complexity: O(h + m), where m is number of k values
    """
    k_set = set(k_values)
    max_k = max(k_values)
    results = {}
    count = [0]  # Use list to modify in nested function
    
    def reverse_inorder(node):
        if node is None or count[0] >= max_k:
            return
        
        # Traverse right subtree
        reverse_inorder(node.right)
        
        # Visit current node
        count[0] += 1
        if count[0] in k_set:
            results[count[0]] = node.val
        
        if count[0] >= max_k:
            return
        
        # Traverse left subtree
        reverse_inorder(node.left)
    
    reverse_inorder(root)
    return results

# Test
result = kthLargestMultiple(root, [1, 3, 5])
print(result)  # {1: 40, 3: 15, 5: 4}

🎓 Interview Key Points Summary

Key Knowledge Points

✅ BST Properties - In-order gives ascending, reverse in-order gives descending
✅ Recursive Traversal - Understand recursion stack depth
✅ Problem Clarification - Edge cases, error handling, data structure definition
✅ Optimization - Extend from single k to multiple k values

Follow-up Questions

🚀 VO Interview Strategy

VO Interview

📚 Related Resources

Official Resources

Learning Resources

Community Discussion

💼 Need Professional Assistance?

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✅ VO Support - Real-time hints and code suggestions
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✅ System Design Coaching - Help you understand big tech thinking
✅ Mock Interview - Practice before the real interview

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Published: March 16, 2026
Difficulty: ⭐⭐⭐⭐ (Medium-Hard)
Tags: Uber, VO Assistance, VO Support, Interview Proxy, Binary Search Tree, Reverse In-Order Traversal, Tree DP, System Design

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