BST Inorder Predecessor in DSA Javascript

Mental Model

The inorder predecessor of a node in a BST is the node that comes just before it when we list all nodes in order from smallest to largest.

Analogy: Imagine a line of people sorted by height. The inorder predecessor of a person is the person just shorter than them standing immediately before in line.

      5
     / \
    3   7
   / \   \
  2   4   8
 ↑curr

Dry Run Walkthrough

Input: BST: 5 -> 3 -> 7 -> 2 -> 4 -> 8, find inorder predecessor of node with value 5

Goal: Find the node that comes just before 5 in the sorted order of BST nodes

Step 1: Start at node 5, check if left child exists

      5
     / \
    3   7
   / \   \
  2   4   8
 ↑curr

Why: We look left because predecessor is largest node smaller than current

Step 2: Move to left child 3, then go right as far as possible

      5
     / \
    3   7
   / \   \
  2   4   8
       ↑curr

Why: Rightmost node in left subtree is the predecessor

Step 3: Move right from 3 to 4, no further right child

      5
     / \
    3   7
   / \   \
  2   4   8
           ↑curr

Why: 4 is the largest node smaller than 5, so it is the predecessor

Result:

      5
     / \
    3   7
   / \   \
  2   4   8
           ↑curr
Inorder predecessor of 5 is 4

Annotated Code

DSA Javascript

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

function findInorderPredecessor(root, targetVal) {
  // Find the node with targetVal
  let curr = root;
  while (curr !== null && curr.val !== targetVal) {
    if (targetVal < curr.val) {
      curr = curr.left; // move left to find target
    } else {
      curr = curr.right; // move right to find target
    }
  }
  if (curr === null) return null; // target not found

  // If left subtree exists, predecessor is rightmost node there
  if (curr.left !== null) {
    let pred = curr.left;
    while (pred.right !== null) {
      pred = pred.right; // move right to find max in left subtree
    }
    return pred.val;
  }

  // If no left subtree, predecessor is ancestor where current node is in right subtree
  let predecessor = null;
  let ancestor = root;
  while (ancestor !== curr) {
    if (curr.val > ancestor.val) {
      predecessor = ancestor; // potential predecessor
      ancestor = ancestor.right; // move right
    } else {
      ancestor = ancestor.left; // move left
    }
  }
  return predecessor ? predecessor.val : null;
}

// Build example BST
const root = new Node(5);
root.left = new Node(3);
root.right = new Node(7);
root.left.left = new Node(2);
root.left.right = new Node(4);
root.right.right = new Node(8);

const target = 5;
const pred = findInorderPredecessor(root, target);
console.log(`Inorder predecessor of ${target} is ${pred}`);

while (curr !== null && curr.val !== targetVal) {

search for the node with targetVal in BST

if (curr.left !== null) {

if left subtree exists, find rightmost node there

while (pred.right !== null) {

move right to find max node in left subtree

while (ancestor !== curr) {

if no left subtree, find ancestor predecessor

if (curr.val > ancestor.val) {

update predecessor when moving right in ancestor path

Output

Inorder predecessor of 5 is 4

Complexity Analysis

Time: O(h) because we traverse from root to target and possibly down one subtree, where h is tree height

Space: O(1) because we use only a few pointers, no extra data structures

vs Alternative: Compared to inorder traversal of entire tree O(n), this approach is faster as it uses BST properties to find predecessor in O(h)

Edge Cases

Target node has no left subtree and is the smallest node

No predecessor exists, function returns null

DSA Javascript

return predecessor ? predecessor.val : null;

Target node not found in BST

Function returns null indicating no such node

DSA Javascript

if (curr === null) return null;

Target node has left subtree with only one node

That single left child is the predecessor

DSA Javascript

while (pred.right !== null) { pred = pred.right; }

When to Use This Pattern

When asked to find the previous node in sorted order in a BST, reach for the inorder predecessor pattern because it uses BST structure to find it efficiently.

Common Mistakes

Mistake: Trying to find predecessor by full inorder traversal instead of using BST properties
Fix: Use BST rules: predecessor is max node in left subtree or ancestor where current is in right subtree

Mistake: Not handling case when node has no left subtree and predecessor is ancestor
Fix: Add ancestor traversal to find predecessor when left subtree is missing

Summary

Finds the node that comes just before a given node in BST sorted order.

Use when you need the previous smaller value relative to a node in a BST.

The predecessor is the rightmost node in left subtree or the nearest ancestor smaller than the node.

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