Ready to take your data structure skills to the next level? Our free BST practice test is the perfect way to challenge yourself with binary search tree quiz scenarios and sharpen your algorithm prowess while boosting your confidence under timed conditions. Test your BST knowledge and tackle diverse BST quiz questions and binary tree practice problems, all designed to help you master node insertion, traversal, and balancing techniques. Curious for more? Dive into our binary questions collection or refresh core concepts with a searching and sorting algorithms quiz . Jump in now and start tracking your progress toward tree-traversal triumph today!
What property uniquely defines a binary search tree?
Every node has at most two children
There are no duplicate values allowed
The tree is always balanced
All left subtree values are less than the node and all right subtree values are greater than the node
A BST is defined by the ordering property: every node's left descendants have smaller values and right descendants have larger values. This invariant holds recursively for all subtrees, enabling efficient search. Other trees may have two children or disallow duplicates but lack this ordering guarantee. Learn more.
What is the average-case time complexity for search operation in a balanced BST?
O(log n)
O(n log n)
O(1)
O(n)
In a balanced BST, the height is proportional to log?(n), so searching down from the root to a leaf takes O(log n) time. Unbalanced trees can degrade to O(n) in the worst case. The log factor comes from halving the search space at each level. Reference.
Which traversal of a BST yields elements in ascending order?
Preorder traversal
Level-order traversal
Inorder traversal
Postorder traversal
Inorder traversal visits the left subtree, root, then right subtree in sequence, which on a BST produces sorted values. Preorder and postorder do not guarantee sorted output. Level-order visits by breadth first, not value order. Details.
After inserting the values [5, 2, 7, 1, 3] into an empty BST, what is the left child of the root?
1
3
2
7
Starting with 5 as root, inserting 2 places it to the left of 5. Subsequent inserts (7 to the right of 5, 1 to left of 2, 3 to right of 2) do not change the left child of 5. Hence the root's left child remains 2. More info.
When deleting a node with two children in a BST, which node is commonly used to replace it?
Preorder predecessor
Inorder successor
Any leaf node
Leftmost child in the left subtree
The standard deletion algorithm replaces a node with two children by its inorder successor (the smallest node in its right subtree). This preserves the BST ordering invariant. Alternatively, the inorder predecessor (largest in left subtree) can be used. See details.
What is the height of a perfectly balanced BST containing 15 nodes (counting edges from root to leaf)?
7
3
4
15
A perfectly balanced BST of height h has 2^(h+1)?1 nodes. Solving 2^(h+1)?1=15 gives h=3. At height 3 there are 4 levels (0 to 3) and 15 total nodes. Learn more.
Which algorithm can validate if a binary tree is a BST in O(n) time using O(1) extra space?
Recursive min/max checks at each node
Morris inorder traversal with a prev pointer
Level-order traversal comparing children
Collecting nodes and sorting then comparing
Morris inorder traversal uses no extra stack or recursion and visits nodes in-order while checking current value against a previous value pointer. This verifies BST order in O(n) time and O(1) space. Recursive methods use O(h) space. Reference.
In a BST that has become a linked list due to ordered inserts, what is the time complexity of a search operation?
O(n)
O(n log n)
O(log n)
O(1)
If a BST degenerates into a linked list (every node has only one child), operations like search degrade to linear time O(n), since each comparison moves only one node further. Balanced BSTs avoid this worst case. More info.
How many structurally unique BSTs can be formed with 4 distinct nodes?
5
42
132
14
The number of unique BSTs for n nodes is the nth Catalan number. For n=4, C? = (1/(4+1))·(8 choose 4) = 14. This counts all distinct structures. Catalan numbers.
What is the worst-case time complexity of building a BST by inserting n sorted values in order?
O(n)
O(n log n)
O(log n)
O(n^2)
Inserting sorted data into an empty BST causes each insert to traverse all existing nodes, resulting in O(1) + O(2) + … + O(n) = O(n²) time. Balanced structures avoid this. Read more.
What is the recurrence for the minimum number of nodes N(h) in an AVL tree of height h?
N(h) = h + 1
N(h) = 2^h ? 1
N(h) = h*(h+1)/2
N(h) = N(h?1) + N(h?2) + 1
An AVL tree maintains balance so the minimum nodes at height h satisfy N(h) = 1 + N(h?1) + N(h?2). This Fibonacci-like recurrence ensures height remains logarithmic. AVL details.
Which data augmentation allows finding the kth smallest element in a BST in O(h) time?
Recording height of each subtree
Keeping pointers to parent nodes
Maintaining cumulative key sums
Storing subtree node counts at each node
By augmenting each node with the size of its left subtree, you can decide whether to go left, right, or return the current node when searching for the kth smallest. This runs in O(h). See example.
In an AVL tree, after inserting a node into the left subtree of the right child of an unbalanced node, which rotation is needed to rebalance?
Left-Right rotation
Single left rotation
Single right rotation
Right-Left rotation
This scenario is a Right-Left case: the heavy side is right, but its left child is heavier. To rebalance, first rotate the right child right, then rotate the unbalanced node left (a double RL rotation). Rotation cases.
What is the main advantage of constructing a BST from a sorted array by choosing the middle element as the root at each step?
It ensures minimal height guaranteeing O(log n) operations
It minimizes the number of leaf nodes
It avoids the need for any rotations
It maximizes the balance factor for each node
By always picking the middle, you build a perfectly balanced BST with minimal height, ensuring search, insert, and delete operations remain O(log n). While rotations aren't used, the key benefit is the height property. Method overview.
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AI Study Notes
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Study Outcomes
Understand BST fundamentals -
Grasp the structure and properties of nodes in a binary search tree to confidently tackle BST practice test questions.
Apply traversal algorithms -
Perform in-order, pre-order, and post-order traversals to retrieve correct node sequences in the binary search tree quiz scenarios.
Construct and modify BSTs -
Insert and delete nodes while maintaining the binary search tree property during interactive practice problems.
Analyze time complexities -
Evaluate the best-case and worst-case runtimes for search, insertion, and deletion operations to optimize BST algorithm performance.
Track progress -
Monitor quiz results to identify strengths and weaknesses, guiding targeted practice for improved BST knowledge.
Cheat Sheet
BST Fundamental Ordering Property -
Binary search trees maintain the invariant that all nodes in the left subtree are less than the parent and all nodes in the right subtree are greater. This property guarantees that an in-order traversal outputs a sorted sequence (CLRS, Chapter 12). Mnemonic: "Left Is Less, Right Is Rest."
Traversal Techniques -
Master in-order, pre-order, and post-order traversals: in-order yields sorted order, pre-order is great for tree copying, and post-order is used in deletion routines (MIT OpenCourseWare, 6.006). For example, an in-order on [5, 3, 7, 2, 4] gives [2, 3, 4, 5, 7].
Insertion & Deletion Operations -
Insertion involves descending to a null child spot and creating a new node (Stanford CS Library). Deletion has three cases: leaf removal, single-child bypass, and two-child replacement via in-order successor or predecessor. Remember: "Find the successor, swap, and pop."
Tree Height & Balance -
Unbalanced BSTs degrade to O(n) operations, while balanced trees (e.g., AVL, red”black) maintain O(log n) height (Journal of Algorithms, 2005). The minimum number of nodes in an AVL tree of height h follows N(h)=N(h-1)+N(h-2)+1, similar to Fibonacci. Aim for |height(left)−height(right)| ≤1.
Practice with BST Quiz Questions -
Regularly tackle BST practice test problems on platforms like LeetCode and HackerRank to solidify concepts (ACM ICPC archives). Mix easy insert/traverse puzzles with harder deletion and balancing tasks to track your BST knowledge growth. Tip: log time and success rate to boost confidence!
