BST Search Operation in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of searching for 7 in this BST?

Consider the following JavaScript code that builds a simple Binary Search Tree (BST) and searches for the value 7. What will be the output printed?

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

class BST {
  constructor() {
    this.root = null;
  }

  insert(value) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          return;
        }
        current = current.left;
      } else {
        if (!current.right) {
          current.right = newNode;
          return;
        }
        current = current.right;
      }
    }
  }

  search(value) {
    let current = this.root;
    while (current) {
      if (value === current.value) {
        return true;
      } else if (value < current.value) {
        current = current.left;
      } else {
        current = current.right;
      }
    }
    return false;
  }
}

const bst = new BST();
bst.insert(10);
bst.insert(5);
bst.insert(15);
bst.insert(3);
bst.insert(7);
bst.insert(12);
bst.insert(18);

console.log(bst.search(7));

A false

B true

C undefined

D TypeError

💡 Hint

Trace the path from root 10 to find 7 by comparing values.

The search starts at the root (10). Since 7 is less than 10, it moves left to 5. Then 7 is greater than 5, so it moves right to 7. It finds the value and returns true.

❓ Predict Output

intermediate

What is the output when searching for 6 in this BST?

Given the same BST as before, what will be the output of searching for the value 6?

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

class BST {
  constructor() {
    this.root = null;
  }

  insert(value) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          return;
        }
        current = current.left;
      } else {
        if (!current.right) {
          current.right = newNode;
          return;
        }
        current = current.right;
      }
    }
  }

  search(value) {
    let current = this.root;
    while (current) {
      if (value === current.value) {
        return true;
      } else if (value < current.value) {
        current = current.left;
      } else {
        current = current.right;
      }
    }
    return false;
  }
}

const bst = new BST();
bst.insert(10);
bst.insert(5);
bst.insert(15);
bst.insert(3);
bst.insert(7);
bst.insert(12);
bst.insert(18);

console.log(bst.search(6));

A false

B true

C null

D ReferenceError

💡 Hint

Check if 6 exists by moving left or right from root.

The search moves from 10 to 5 (since 6 < 10), then from 5 to 7 (since 6 > 5). Since 7 has no left child, the search ends without finding 6, returning false.

🔧 Debug

advanced

What error does this BST search code produce?

Examine the following BST search method. What error will it cause when searching for any value?

DSA Javascript

search(value) {
  let current = this.root;
  while (current !== null) {
    if (value = current.value) {
      return true;
    } else if (value < current.value) {
      current = current.left;
    } else {
      current = current.right;
    }
  }
  return false;
}

A No error, returns correct result

B TypeError

C Infinite loop

D SyntaxError

💡 Hint

Look carefully at the if condition inside the while loop.

The condition uses assignment (=) instead of comparison (===). This assigns current.value to value, which is truthy, so the if condition is always true, causing an infinite loop.

🧠 Conceptual

advanced

How many nodes are visited when searching for 12 in this BST?

Given the BST constructed by inserting values in this order: 10, 5, 15, 3, 7, 12, 18. How many nodes will the search method visit when searching for 12?

A 3

B 4

C 5

D 2

💡 Hint

Trace the path from root to 12 counting nodes visited.

Search path: 10 (root) → 15 (right child) → 12 (left child of 15). Total nodes visited = 3.

❓ Predict Output

expert

What is the output of searching for 20 in this BST after these operations?

After inserting 10, 5, 15, 3, 7, 12, 18 into a BST, the following code runs. What is the output?

DSA Javascript

class Node {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;
  }
}

class BST {
  constructor() {
    this.root = null;
  }

  insert(value) {
    const newNode = new Node(value);
    if (!this.root) {
      this.root = newNode;
      return;
    }
    let current = this.root;
    while (true) {
      if (value < current.value) {
        if (!current.left) {
          current.left = newNode;
          return;
        }
        current = current.left;
      } else {
        if (!current.right) {
          current.right = newNode;
          return;
        }
        current = current.right;
      }
    }
  }

  search(value) {
    let current = this.root;
    while (current) {
      if (value === current.value) {
        return true;
      } else if (value < current.value) {
        current = current.left;
      } else {
        current = current.right;
      }
    }
    return false;
  }
}

const bst = new BST();
bst.insert(10);
bst.insert(5);
bst.insert(15);
bst.insert(3);
bst.insert(7);
bst.insert(12);
bst.insert(18);

console.log(bst.search(20));

A RangeError

B true

C undefined

D false

💡 Hint

20 is greater than all inserted values; check rightmost path.

Search path: 10 → 15 → 18. Since 20 > 18 and 18 has no right child, search ends returning false.

Practice

1. What is the main purpose of the search operation in a Binary Search Tree (BST)?

easy

A. To insert a new value into the BST

B. To find whether a given value exists in the BST

C. To delete a value from the BST

D. To balance the BST

Solution

1. Step 1: Understand the search operation

   The search operation in a BST is used to check if a specific value is present in the tree.

2. Step 2: Differentiate from other operations

   Insertion adds new values, deletion removes values, and balancing adjusts the tree structure. Search only checks for existence.

3. Final Answer:

   To find whether a given value exists in the BST -> Option B

4. Quick Check:

   Search = Find value [OK]

Hint: Search checks presence, not insertion or deletion [OK]

Common Mistakes:

• Confusing search with insert or delete
• Thinking search changes the tree
• Assuming search balances the tree

2. Which of the following is the correct syntax to start a search function for a BST node in JavaScript?

easy

A. function searchBST(value, root) { }

B. function searchBST(value) { return root; }

C. function searchBST(root) { return value; }

D. function searchBST(root, value) { }

Solution

1. Step 1: Identify parameters needed for BST search

   The search needs the root node to start and the value to find, so parameters are (root, value).

2. Step 2: Check function signature correctness

   function searchBST(root, value) { } correctly uses (root, value). Options A, C, and D have wrong parameter order or missing parameters.

3. Final Answer:

   function searchBST(root, value) { } -> Option D

4. Quick Check:

   Correct parameters order = function searchBST(root, value) { } [OK]

Hint: Root node first, value second in search function [OK]

Common Mistakes:

• Swapping parameter order
• Missing one parameter
• Returning wrong variable immediately

3. Consider the following BST and the search code snippet:

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

const root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);

function searchBST(root, val) {
  if (!root) return false;
  if (root.val === val) return true;
  if (val < root.val) return searchBST(root.left, val);
  else return searchBST(root.right, val);
}

console.log(searchBST(root, 15));

What will be the output?

medium

A. true

B. false

C. undefined

D. Error

Solution

1. Step 1: Trace the search for value 15

   Start at root (10). Since 15 > 10, move to right child (15).

2. Step 2: Check if current node value matches 15

   Node value is 15, which matches the search value, so return true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Value found = true [OK]

Hint: Compare value with node, go left or right [OK]

Common Mistakes:

• Returning false too early
• Not checking node value before recursion
• Confusing left and right child directions

4. The following code is intended to search a value in a BST but does not work correctly. What is the error?

function searchBST(root, val) {
  if (root === null) return false;
  if (root.val === val) return true;
  if (val > root.val) return searchBST(root.left, val);
  else return searchBST(root.right, val);
}

medium

A. Incorrect return type

B. Missing base case for empty tree

C. The comparison directions for left and right child are swapped

D. Function parameters are wrong

Solution

1. Step 1: Analyze comparison logic for recursion

   If val > root.val, search should go right, but code goes left. This is reversed.

2. Step 2: Confirm correct child direction for BST search

   For val > root.val, search right subtree; for val < root.val, search left subtree.

3. Final Answer:

   The comparison directions for left and right child are swapped -> Option C

4. Quick Check:

   Left/right child direction swapped = The comparison directions for left and right child are swapped [OK]

Hint: Greater value goes right, smaller goes left [OK]

Common Mistakes:

• Swapping left and right child in recursion
• Forgetting base case
• Returning wrong data type

5. You have a BST with nodes: 20 (root), 10 (left child), 30 (right child), 5 (left of 10), 15 (right of 10). You want to search for value 12. Which nodes will the search visit in order?

hard

A. 20 -> 10 -> 15

B. 20 -> 30 -> 15

C. 20 -> 10 -> 5

D. 20 -> 15 -> 10

Solution

1. Step 1: Start at root node 20 and compare with 12

   Since 12 < 20, move to left child 10.

2. Step 2: Compare 12 with node 10

   12 > 10, so move to right child 15.

3. Step 3: Compare 12 with node 15

   12 < 15, so move to left child, which is null (not in tree), search ends.

4. Final Answer:

   20 -> 10 -> 15 -> Option A

5. Quick Check:

   Visited nodes = 20, 10, 15 [OK]

Hint: Go left if smaller, right if bigger, stop if null [OK]

Common Mistakes:

• Visiting wrong child nodes
• Assuming 12 is in the tree
• Skipping nodes in path