BST Delete Operation in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after deleting a leaf node in BST

What is the printed state of the BST after deleting the leaf node with value 20?

DSA Javascript

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

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

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

  delete(data) {
    this.root = this._deleteRec(this.root, data);
  }

  _deleteRec(root, data) {
    if (!root) return root;
    if (data < root.data) {
      root.left = this._deleteRec(root.left, data);
    } else if (data > root.data) {
      root.right = this._deleteRec(root.right, data);
    } else {
      if (!root.left && !root.right) {
        return null;
      } else if (!root.left) {
        return root.right;
      } else if (!root.right) {
        return root.left;
      } else {
        let minLargerNode = this._minValueNode(root.right);
        root.data = minLargerNode.data;
        root.right = this._deleteRec(root.right, minLargerNode.data);
      }
    }
    return root;
  }

  _minValueNode(node) {
    let current = node;
    while (current.left) {
      current = current.left;
    }
    return current;
  }

  inorder(root = this.root, result = []) {
    if (root) {
      this.inorder(root.left, result);
      result.push(root.data);
      this.inorder(root.right, result);
    }
    return result;
  }

  print() {
    const values = this.inorder();
    console.log(values.join(' -> ') + ' -> null');
  }
}

const bst = new BST();
[50, 30, 70, 20, 40, 60, 80].forEach(n => bst.insert(n));
bst.delete(20);
bst.print();

A 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> null

B 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> null

C 50 -> 30 -> 40 -> 60 -> 70 -> 80 -> null

D 30 -> 40 -> 50 -> 60 -> 70 -> null

💡 Hint

Deleting a leaf node removes it without affecting other nodes.

Deleting 20 removes the leaf node 20. The inorder traversal then lists nodes in ascending order without 20.

❓ Predict Output

intermediate

Output after deleting a node with one child in BST

What is the printed state of the BST after deleting the node with value 30, which has one child?

DSA Javascript

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

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

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

  delete(data) {
    this.root = this._deleteRec(this.root, data);
  }

  _deleteRec(root, data) {
    if (!root) return root;
    if (data < root.data) {
      root.left = this._deleteRec(root.left, data);
    } else if (data > root.data) {
      root.right = this._deleteRec(root.right, data);
    } else {
      if (!root.left && !root.right) {
        return null;
      } else if (!root.left) {
        return root.right;
      } else if (!root.right) {
        return root.left;
      } else {
        let minLargerNode = this._minValueNode(root.right);
        root.data = minLargerNode.data;
        root.right = this._deleteRec(root.right, minLargerNode.data);
      }
    }
    return root;
  }

  _minValueNode(node) {
    let current = node;
    while (current.left) {
      current = current.left;
    }
    return current;
  }

  inorder(root = this.root, result = []) {
    if (root) {
      this.inorder(root.left, result);
      result.push(root.data);
      this.inorder(root.right, result);
    }
    return result;
  }

  print() {
    const values = this.inorder();
    console.log(values.join(' -> ') + ' -> null');
  }
}

const bst = new BST();
[50, 30, 70, 40, 60, 80].forEach(n => bst.insert(n));
bst.delete(30);
bst.print();

A 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> null

B 40 -> 50 -> 60 -> 70 -> 80 -> null

C 50 -> 40 -> 60 -> 70 -> 80 -> null

D 40 -> 30 -> 50 -> 60 -> 70 -> 80 -> null

💡 Hint

When deleting a node with one child, the child replaces the node.

Node 30 has one child 40. After deletion, 40 replaces 30. The inorder traversal is [40, 50, 60, 70, 80].

❓ Predict Output

advanced

Output after deleting a node with two children in BST

What is the printed state of the BST after deleting the node with value 50, which has two children?

DSA Javascript

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

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

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

  delete(data) {
    this.root = this._deleteRec(this.root, data);
  }

  _deleteRec(root, data) {
    if (!root) return root;
    if (data < root.data) {
      root.left = this._deleteRec(root.left, data);
    } else if (data > root.data) {
      root.right = this._deleteRec(root.right, data);
    } else {
      if (!root.left && !root.right) {
        return null;
      } else if (!root.left) {
        return root.right;
      } else if (!root.right) {
        return root.left;
      } else {
        let minLargerNode = this._minValueNode(root.right);
        root.data = minLargerNode.data;
        root.right = this._deleteRec(root.right, minLargerNode.data);
      }
    }
    return root;
  }

  _minValueNode(node) {
    let current = node;
    while (current.left) {
      current = current.left;
    }
    return current;
  }

  inorder(root = this.root, result = []) {
    if (root) {
      this.inorder(root.left, result);
      result.push(root.data);
      this.inorder(root.right, result);
    }
    return result;
  }

  print() {
    const values = this.inorder();
    console.log(values.join(' -> ') + ' -> null');
  }
}

const bst = new BST();
[50, 30, 70, 20, 40, 60, 80].forEach(n => bst.insert(n));
bst.delete(50);
bst.print();

A 20 -> 30 -> 40 -> 60 -> 70 -> 80 -> null

B 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> null

C 20 -> 30 -> 40 -> 60 -> 70 -> 80 -> 50 -> null

D 30 -> 40 -> 60 -> 70 -> 80 -> null

💡 Hint

When deleting a node with two children, replace it with the smallest node in the right subtree.

Node 50 is replaced by 60 (smallest in right subtree). Then 60 is deleted from its original position. The inorder traversal is [20, 30, 40, 60, 70, 80].

🧠 Conceptual

advanced

Understanding BST delete operation cases

Which of the following correctly describes the three main cases when deleting a node from a BST?

A Node is left child; node is right child; node is root

B Node is root; node is leaf; node is internal

C Node has no children; node has left child only; node has right child only

D Node is a leaf; node has one child; node has two children

💡 Hint

Think about how many children the node to delete can have.

The three main cases are deleting a leaf node (no children), a node with one child, and a node with two children.

🔧 Debug

expert

Identify the bug in BST delete operation code

Given the following snippet from a BST delete function, what is the bug causing incorrect deletion when the node has two children? ```js _deleteRec(root, data) { if (!root) return root; if (data < root.data) { root.left = this._deleteRec(root.left, data); } else if (data > root.data) { root.right = this._deleteRec(root.right, data); } else { if (!root.left && !root.right) { return null; } else if (!root.left) { return root.right; } else if (!root.right) { return root.left; } else { let minLargerNode = this._minValueNode(root.right); root.data = minLargerNode.data; root.left = this._deleteRec(root.right, minLargerNode.data); } } return root; } ``` What is the bug?

A The base case should return null instead of root

B The function does not handle leaf nodes correctly

C The recursive call after replacing data should update root.right, not root.left

D The minimum value node is found incorrectly from the left subtree

💡 Hint

Check which subtree is updated after replacing the node's data.

After replacing the node's data with the minimum from the right subtree, the deletion must continue in the right subtree. The code incorrectly updates root.left instead of root.right.

Practice

1. What happens when you delete a leaf node (a node with no children) from a Binary Search Tree (BST)?

easy

A. The node is simply removed without affecting other nodes.

B. The entire tree is rebuilt from scratch.

C. The node's children replace it in the tree.

D. The root node is always changed.

Solution

1. Step 1: Understand leaf node deletion

   A leaf node has no children, so removing it does not affect other nodes.

2. Step 2: Effect on BST structure

   Since no children exist, simply remove the node and update its parent's pointer to null.

3. Final Answer:

   The node is simply removed without affecting other nodes. -> Option A

4. Quick Check:

   Leaf node deletion = simple removal [OK]

Hint: Leaf nodes delete by direct removal only [OK]

Common Mistakes:

• Thinking children replace the node when none exist
• Assuming root always changes
• Believing tree rebuild is needed

2. Which of the following is the correct syntax to delete a node with value key from a BST in JavaScript?

easy

A. bst.deleteNode(key);

B. bst.delete(key);

C. bst.remove(key);

D. bst.erase(key);

Solution

1. Step 1: Identify common BST delete method

   In many JavaScript BST implementations, the method to delete a node is named delete.

2. Step 2: Check method naming conventions

   Methods like remove, erase, or deleteNode are not standard or common in JavaScript BST examples.

3. Final Answer:

   bst.delete(key); -> Option B

4. Quick Check:

   Standard BST delete method = delete() [OK]

Hint: Use bst.delete(key) to remove nodes [OK]

Common Mistakes:

• Using remove() or erase() which are not standard
• Confusing delete keyword with method name
• Using deleteNode() which is uncommon

3. Consider this BST and the delete operation:

    10
   /  \
  5    15
       / \
      12  20

What will be the in-order traversal output after deleting node with value 15?

medium

A. [5, 10, 12, 15, 20]

B. [5, 10, 20]

C. [5, 10, 12, 20]

D. [5, 10, 15, 20]

Solution

1. Step 1: Identify node to delete and its children

   Node 15 has two children: 12 (left) and 20 (right).

2. Step 2: Replace node 15 with its in-order successor

   The in-order successor is the smallest node in right subtree, which is 20. Replace 15 with 20 and delete original 20 node.

3. Step 3: Perform in-order traversal

   In-order traversal visits nodes in ascending order: 5, 10, 12, 20.

4. Final Answer:

   [5, 10, 12, 20] -> Option C

5. Quick Check:

   In-order after deleting 15 = [5, 10, 12, 20] [OK]

Hint: Replace deleted node with in-order successor [OK]

Common Mistakes:

• Removing 15 without replacement
• Using in-order predecessor instead of successor
• Forgetting to delete the successor node

4. The following code attempts to delete a node from a BST but causes an error. What is the main issue?

function deleteNode(root, key) {
  if (!root) return null;
  if (key < root.val) root.left = deleteNode(root.left, key);
  else if (key > root.val) root.right = deleteNode(root.right, key);
  else {
    if (!root.left) return root.right;
    else if (!root.right) return root.left;
    let minNode = findMin(root.right);
    root.val = minNode.val;
    // Missing deletion of minNode here
  }
  return root;
}

function findMin(node) {
  while (node.left) node = node.left;
  return node;
}

medium

A. The function does not handle leaf nodes correctly.

B. The base case for recursion is missing.

C. The function uses incorrect comparison operators.

D. The code does not delete the in-order successor node after copying its value.

Solution

1. Step 1: Analyze deletion of node with two children

   When deleting a node with two children, we replace its value with the in-order successor's value.

2. Step 2: Check if in-order successor node is deleted

   The code copies the successor's value but does not delete the successor node itself, causing duplicate values and incorrect tree.

3. Final Answer:

   The code does not delete the in-order successor node after copying its value. -> Option D

4. Quick Check:

   Must delete successor node after value copy [OK]

Hint: Always delete successor node after copying value [OK]

Common Mistakes:

• Forgetting to delete the in-order successor node
• Missing base case (though present here)
• Incorrectly handling leaf nodes
• Using wrong comparison operators

5. You have a BST with nodes: 8, 3, 10, 1, 6, 14, 4, 7, 13.
After deleting node 3, what is the correct in-order traversal of the BST?

hard

A. [1, 4, 6, 7, 8, 10, 13, 14]

B. [1, 3, 4, 6, 7, 8, 10, 13, 14]

C. [1, 6, 7, 8, 10, 13, 14]

D. [1, 4, 7, 8, 10, 13, 14]

Solution

1. Step 1: Identify node 3 and its children

   Node 3 has two children: 1 (left) and 6 (right). 6 has children 4 and 7.

2. Step 2: Delete node 3 by replacing with in-order successor

   The in-order successor of 3 is 4 (smallest in right subtree). Replace 3 with 4, then delete original 4 node.

3. Step 3: Perform in-order traversal after deletion

   In-order traversal visits nodes in ascending order: 1, 4, 6, 7, 8, 10, 13, 14.

4. Final Answer:

   [1, 4, 6, 7, 8, 10, 13, 14] -> Option A

5. Quick Check:

   In-order after deleting 3 = [1, 4, 6, 7, 8, 10, 13, 14] [OK]

Hint: Replace deleted node with in-order successor, then traverse [OK]

Common Mistakes:

• Forgetting to delete the in-order successor node
• Using in-order predecessor instead
• Skipping nodes in traversal