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.