BST Inorder Predecessor in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

function getLeftChild(node) {
  return node.[1];
}

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

Aright

Bleft

Cparent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

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

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

2fill in blank

medium

Complete the code to move to the right child of a node in a BST.

DSA Javascript

function getRightChild(node) {
  return node.[1];
}

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

Aright

Bparent

Cleft

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

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

The right child of a BST node is accessed using the 'right' property.

3fill in blank

hard

Fix the error in the code to find the inorder predecessor when the left subtree exists.

DSA Javascript

function inorderPredecessor(node) {
  let current = node.[1];
  while (current.right !== null) {
    current = current.right;
  }
  return current;
}

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

Aright

Bvalue

Cleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from node.right instead of node.left.

To find the inorder predecessor, start from the left child and go right as far as possible.

4fill in blank

hard

Fill both blanks to check if a node has a left child and then find its rightmost child.

DSA Javascript

if (node.[1] !== null) {
  let current = node.[2];
  while (current.right !== null) {
    current = current.right;
  }
  return current;
}

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

Aleft

Bright

Cparent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using different properties for check and traversal.

Check if node.left exists, then start from node.left to find the rightmost node.

5fill in blank

hard

Fill all three blanks to complete the function that finds the inorder predecessor when no left child exists.

DSA Javascript

function inorderPredecessor(node) {
  if (node.left !== null) {
    let current = node.left;
    while (current.right !== null) {
      current = current.right;
    }
    return current;
  } else {
    let current = node;
    let parent = node.[1];
    while (parent !== null && current === parent.[2]) {
      current = parent;
      parent = parent.[3];
    }
    return parent;
  }
}

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

Aleft

Bright

Cparent

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' instead of 'right' in the while condition.

Not moving up the parent chain correctly.

When no left child, move up using parent pointers until current is not the right child of parent.

Practice

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

easy

A. The node with the largest value smaller than the given node

B. The node with the smallest value larger than the given node

C. The root node of the BST

D. The left child of the given node

Solution

1. Step 1: Understand BST ordering

   In a BST, left children are smaller and right children are larger than the node.

2. Step 2: Define inorder predecessor

   The inorder predecessor is the node visited just before the current node in an inorder traversal, which means it has the largest value smaller than the current node.

3. Final Answer:

   The node with the largest value smaller than the given node -> Option A

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Inorder predecessor is largest smaller value [OK]

Common Mistakes:

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

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

easy

A. function findPredecessor(node) {}

B. function inorderSuccessor(root, node) {}

C. function findInorderPredecessor(root, node) {}

D. function getInorderPredecessor(node, root) {}

Solution

1. Step 1: Check function purpose and parameters

   The function should find the inorder predecessor given the BST root and the target node.

2. Step 2: Match signature to purpose

   function findInorderPredecessor(root, node) {} correctly uses root and node as parameters and names the function appropriately.

3. Final Answer:

   function findInorderPredecessor(root, node) {} -> Option C

4. Quick Check:

   Correct function name and parameters [OK]

Hint: Function needs root and node parameters [OK]

Common Mistakes:

• Using successor function name instead of predecessor
• Swapping parameter order
• Missing root parameter

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

       20
      /  \
    10    30
      \     \
      15     40

medium

A. 10

B. 15

C. 20

D. 30

Solution

1. Step 1: Locate node 15 and check left subtree

   Node 15 has no left child, so we look upwards to find the largest smaller ancestor.

2. Step 2: Find largest smaller ancestor

   Node 10 is the parent of 15 and smaller than 15, so 10 is the inorder predecessor.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Largest smaller value than 15 is 10 [OK]

Hint: If no left child, go up to largest smaller ancestor [OK]

Common Mistakes:

• Choosing the node itself as predecessor
• Choosing the parent if it is larger
• Confusing successor with predecessor

4. The following JavaScript code tries to find the inorder predecessor of a node in a BST but has a bug. What is the bug?

function findInorderPredecessor(root, node) {
  if (node.left) {
    let temp = node.left;
    while (temp.right) {
      temp = temp.right;
    }
    return temp;
  }
  let predecessor = null;
  let current = root;
  while (current) {
    if (node.val > current.val) {
      predecessor = current;
      current = current.right;
    } else {
      current = current.left;
    }
  }
  return predecessor;
}

medium

A. It does not handle the case when node has no left child

B. It does not check if node exists in the BST

C. It should return null if predecessor is not found

D. It incorrectly updates predecessor when node.val <= current.val

Solution

1. Step 1: Analyze code logic for predecessor

   The code correctly finds predecessor if node has left child or searches ancestors otherwise.

2. Step 2: Identify missing validation

   Code does not verify if the given node actually exists in the BST, which can cause incorrect results or errors.

3. Final Answer:

   It does not check if node exists in the BST -> Option B

4. Quick Check:

   Missing node existence check [OK]

Hint: Always verify node exists in BST before predecessor search [OK]

Common Mistakes:

• Assuming node is always in BST
• Confusing left child check logic
• Ignoring null return possibility

5. You want to find the inorder predecessor of the node with value 25 in the BST below:

          40
         /  \
       20    60
      /  \  /  \
    10   30 50  70
         / 
        25

Which node is the inorder predecessor of 25, and why?

hard

A. 10, because it is the smallest node smaller than 25

B. 30, because it is the parent of 25

C. 25 has no inorder predecessor

D. 20, because it is the largest smaller ancestor

Solution

1. Step 1: Check if node 25 has left child

   Node 25 has no left child, so we look for the largest ancestor smaller than 25.

2. Step 2: Find largest smaller ancestor

   Ancestors of 25 are 30, 20, and 40. Among these, 30 is the parent of 25 and the largest smaller ancestor is 20, but since 25 is in the left subtree of 30, the inorder predecessor is 30's left subtree's rightmost node or the parent if no left subtree exists. Here, 25 is left child of 30, so predecessor is 20.

3. Final Answer:

   20, because it is the largest smaller ancestor -> Option D

4. Quick Check:

   Largest smaller ancestor = 20 [OK]

Hint: If no left child, predecessor is largest smaller ancestor [OK]

Common Mistakes:

• Choosing parent node 30 incorrectly
• Thinking no predecessor exists
• Choosing smallest smaller node instead of largest