BST Inorder Predecessor in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the inorder predecessor of a node in BST

What is the output of the code when finding the inorder predecessor of node with value 15 in the given BST?

DSA Javascript

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

function insert(root, val) {
  if (!root) return new Node(val);
  if (val < root.val) root.left = insert(root.left, val);
  else root.right = insert(root.right, val);
  return root;
}

function inorderPredecessor(root, node) {
  let predecessor = null;
  while (root) {
    if (node.val <= root.val) {
      root = root.left;
    } else {
      predecessor = root;
      root = root.right;
    }
  }
  return predecessor ? predecessor.val : null;
}

const values = [20, 10, 30, 5, 15, 25, 35];
let root = null;
for (const v of values) {
  root = insert(root, v);
}
const node = { val: 15 };
console.log(inorderPredecessor(root, node));

A 5

B 20

C 10

D 15

💡 Hint

Remember, the inorder predecessor is the largest value smaller than the target node.

The inorder predecessor of 15 is 10 because 10 is the largest value less than 15 in the BST.

❓ Predict Output

intermediate

Output of inorder predecessor when node has left subtree

What will be printed when finding the inorder predecessor of node with value 30 in this BST?

DSA Javascript

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

function insert(root, val) {
  if (!root) return new Node(val);
  if (val < root.val) root.left = insert(root.left, val);
  else root.right = insert(root.right, val);
  return root;
}

function findMax(node) {
  while (node.right) {
    node = node.right;
  }
  return node.val;
}

function inorderPredecessor(root, node) {
  if (node.left) {
    return findMax(node.left);
  }
  let predecessor = null;
  while (root) {
    if (node.val <= root.val) {
      root = root.left;
    } else {
      predecessor = root;
      root = root.right;
    }
  }
  return predecessor ? predecessor.val : null;
}

const values = [20, 10, 30, 25, 35];
let root = null;
for (const v of values) {
  root = insert(root, v);
}
const node = { val: 30, left: new Node(25) };
console.log(inorderPredecessor(root, node));

A 20

B 25

C 10

D 35

💡 Hint

If the node has a left subtree, the predecessor is the maximum value in that subtree.

Since node 30 has a left child 25, the inorder predecessor is the maximum value in the left subtree, which is 25.

🔧 Debug

advanced

Identify the error in inorder predecessor function

What error will this code produce when trying to find the inorder predecessor of node with value 10?

DSA Javascript

function inorderPredecessor(root, node) {
  let predecessor = null;
  while (root) {
    if (node.val < root.val) {
      root = root.left;
    } else if (node.val > root.val) {
      predecessor = root;
      root = root.right;
    } else {
      if (root.left) {
        let temp = root.left;
        while (temp.right) {
          temp = temp.right;
        }
        return temp.val;
      }
      break;
    }
  }
  return predecessor ? predecessor.val : null;
}

const root = {
  val: 20,
  left: { val: 10, left: { val: 5, left: null, right: null }, right: { val: 15, left: null, right: null } },
  right: { val: 30, left: null, right: null }
};
const node = { val: 10 };
console.log(inorderPredecessor(root, node));

A 5

B 15

C null

D TypeError

💡 Hint

Check how the code finds the max value in the left subtree.

The inorder predecessor of 10 is 5, found as the rightmost node in the left subtree of 10.

🧠 Conceptual

advanced

Understanding inorder predecessor when no left subtree

If a node in a BST has no left child, how is its inorder predecessor found?

A The rightmost ancestor whose left child is also an ancestor of the node

B The leftmost ancestor whose left child is also an ancestor of the node

C The rightmost ancestor whose right child is also an ancestor of the node

D The leftmost ancestor whose right child is also an ancestor of the node

💡 Hint

Think about moving up the tree to find a smaller value than the node.

When no left subtree exists, the inorder predecessor is the closest ancestor for which the node lies in its right subtree, i.e., the leftmost ancestor whose right child is also an ancestor of the node.

🚀 Application

expert

Find inorder predecessor in BST with parent pointers

Given a BST node with parent pointers, which code snippet correctly finds its inorder predecessor?

A

function inorderPredecessor(node) {
  if (node.left) {
    let curr = node.left;
    while (curr.right) curr = curr.right;
    return curr;
  }
  let curr = node;
  while (curr.parent && curr === curr.parent.left) {
    curr = curr.parent;
  }
  return curr.parent;
}

B

function inorderPredecessor(node) {
  if (node.right) {
    let curr = node.right;
    while (curr.left) curr = curr.left;
    return curr;
  }
  let curr = node;
  while (curr.parent && curr === curr.parent.right) {
    curr = curr.parent;
  }
  return curr.parent;
}

C

function inorderPredecessor(node) {
  if (node.left) {
    let curr = node.left;
    while (curr.left) curr = curr.left;
    return curr;
  }
  let curr = node;
  while (curr.parent && curr === curr.parent.right) {
    curr = curr.parent;
  }
  return curr.parent;
}

D

function inorderPredecessor(node) {
  if (node.left) {
    let curr = node.left;
    while (curr.right) curr = curr.right;
    return curr;
  }
  let curr = node;
  while (curr.parent && curr === curr.parent.right) {
    curr = curr.parent;
  }
  return curr.parent;
}

💡 Hint

Remember the inorder predecessor is the max node in left subtree or the first ancestor where node is in right subtree.

Option A correctly finds the max node in left subtree if exists, else climbs up until node is a right child, then returns that 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