BST Inorder Successor in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the left child of a node in a BST.

DSA Javascript

function leftChild(node) {
  return node[1];
}

Drag options to blanks, or click blank then click option'

A.left

B.right

C.parent

D.child

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.right' instead of '.left' to access the left child.

Using '.parent' which points to the parent node, not the child.

In a BST, the left child of a node is accessed using the '.left' property.

2fill in blank

medium

Complete the code to find the minimum node in a BST subtree.

DSA Javascript

function findMin(node) {
  while (node[1]) {
    node = node.left;
  }
  return node;
}

Drag options to blanks, or click blank then click option'

A.right

B.child

C.parent

D.left

Attempts:

3 left

💡 Hint

Common Mistakes

Checking '.right' instead of '.left' in the loop condition.

Not updating the node inside the loop.

To find the minimum node, keep going left until there is no left child.

3fill in blank

hard

Fix the error in the code to find the inorder successor when the node has no right child.

DSA Javascript

function inorderSuccessor(node) {
  if (node.right) {
    return findMin(node.right);
  }
  let parent = node[1];
  while (parent && node === parent.right) {
    node = parent;
    parent = parent.parent;
  }
  return parent;
}

Drag options to blanks, or click blank then click option'

A.parent

B.left

C.right

D.child

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.left' or '.right' instead of '.parent' to move up the tree.

Not updating the parent inside the loop.

To move up the tree, we use the '.parent' property to find the parent node.

4fill in blank

hard

Fill both blanks to complete the function that finds the inorder successor of a node in a BST.

DSA Javascript

function inorderSuccessor(node) {
  if (node.right) {
    return [1](node.right);
  }
  let parent = node.[2];
  while (parent && node === parent.right) {
    node = parent;
    parent = parent.parent;
  }
  return parent;
}

Drag options to blanks, or click blank then click option'

AfindMin

BfindMax

Cparent

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'findMax' instead of 'findMin' for the right subtree.

Using 'right' or 'left' instead of 'parent' to move up.

If the node has a right child, the successor is the minimum node in the right subtree (findMin). Otherwise, move up using the 'parent' property, which is accessed by 'node.parent'.

5fill in blank

hard

Fill all three blanks to complete the function that returns the inorder successor of a node in a BST.

DSA Javascript

function inorderSuccessor(node) {
  if (node.[1]) {
    return findMin(node.[2]);
  }
  let parent = node.[3];
  while (parent && node === parent.right) {
    node = parent;
    parent = parent.parent;
  }
  return parent;
}

Drag options to blanks, or click blank then click option'

Aright

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using different properties for the first two blanks.

Using 'left' instead of 'parent' to move up.

The function checks if the node has a right child (node.right). If yes, it finds the minimum node in the right subtree (node.right). Otherwise, it moves up the tree using the parent property (node.parent).

Practice

1. What is the inorder successor of a node in a Binary Search Tree (BST)?

easy

A. The node with the smallest key greater than the given node's key

B. The node with the largest key smaller than the given node's key

C. The root node of the BST

D. The left child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order of their keys in a BST.

2. Step 2: Define inorder successor

   The inorder successor of a node is the next node visited after it in inorder traversal, which is the smallest key greater than the node's key.

3. Final Answer:

   The node with the smallest key greater than the given node's key -> Option A

4. Quick Check:

   Inorder successor = next bigger key [OK]

Hint: Inorder successor is next bigger node in sorted order [OK]

Common Mistakes:

• Confusing successor with predecessor
• Thinking successor is always the right child
• Assuming successor is the root node

2. Which of the following is the correct JavaScript function signature to find the inorder successor of a node in a BST?

easy

A. function inorderSuccessor(node) {}

B. function inorderSuccessor(root, node) {}

C. function findSuccessor(root) {}

D. function successor(node, root) {}

Solution

1. Step 1: Identify required inputs

   To find the inorder successor, we need the BST root to traverse and the target node to find its successor.

2. Step 2: Match function signature

   function inorderSuccessor(root, node) {} correctly takes both root and node as parameters, enabling traversal and comparison.

3. Final Answer:

   function inorderSuccessor(root, node) {} -> Option B

4. Quick Check:

   Function needs root and node [OK]

Hint: Successor function needs both root and target node [OK]

Common Mistakes:

• Using only node without root
• Wrong parameter order
• Incorrect function naming

3. Given the BST below and the node with value 15, what is the inorder successor?

      20
     /  \
   10    30
     
     15

medium

A. 10

B. 30

C. 20

D. 15

Solution

1. Step 1: Locate node 15 and its right subtree

   Node 15 has no right child, so successor is an ancestor.

2. Step 2: Find the lowest ancestor whose left child is also ancestor of 15

   Node 20 is the first ancestor where 15 is in left subtree, so successor is 20.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Successor of 15 is 20 [OK]

Hint: If no right child, go up to first bigger ancestor [OK]

Common Mistakes:

• Choosing right child when none exists
• Picking the node itself as successor
• Confusing successor with predecessor

4. Consider this buggy JavaScript code to find the inorder successor of a node in a BST:

function inorderSuccessor(root, node) {
  if (!root) return null;
  if (node.val >= root.val) {
    return inorderSuccessor(root.right, node);
  } else {
    let left = inorderSuccessor(root.left, node);
    return left ? left : root;
  }
}

What is the main issue with this code?

medium

A. It does not handle the case when node has a right subtree

B. It incorrectly compares node.val with root.val

C. It returns the node itself instead of successor

D. It causes infinite recursion

Solution

1. Step 1: Analyze code logic

   The code only traverses the tree comparing values but never checks if node has a right subtree.

2. Step 2: Identify missing right subtree check

   If node has a right child, successor is the leftmost node in right subtree, which this code ignores.

3. Final Answer:

   It does not handle the case when node has a right subtree -> Option A

4. Quick Check:

   Missing right subtree check causes wrong successor [OK]

Hint: Check right subtree first for successor [OK]

Common Mistakes:

• Ignoring right subtree
• Wrong base case
• Returning node instead of successor

5. You have a BST and want to find the inorder successor of node 25. The BST structure is:

        20
       /  \
     10    30
           /  \
         25    40

What is the inorder successor of node 25 and why?

hard

A. 40, because 40 is the right child of 30

B. 20, because it is the root

C. None, because 25 is the largest node in its subtree

D. 30, because 25 has no right child and 30 is the next bigger ancestor

Solution

1. Step 1: Check if node 25 has right subtree

   Node 25 has no right child, so successor is an ancestor.

2. Step 2: Find the lowest ancestor whose left subtree contains node 25

   Node 30 is the parent of 25 and is the next bigger node, so successor is 30.

3. Final Answer:

   30, because 25 has no right child and 30 is the next bigger ancestor -> Option D

4. Quick Check:

   Successor of 25 is 30 [OK]

Hint: If no right child, successor is nearest bigger ancestor [OK]

Common Mistakes:

• Choosing right child of parent instead of parent
• Assuming no successor if no right child
• Confusing root as successor