BST Inorder Predecessor in DSA Javascript - Time & Space Complexity

Time Complexity: BST Inorder Predecessor

O(h)

Understanding Time Complexity

We want to understand how long it takes to find the inorder predecessor in a Binary Search Tree (BST).

Specifically, how the time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function inorderPredecessor(root, node) {
  if (node.left) {
    let pred = node.left;
    while (pred.right) {
      pred = pred.right;
    }
    return pred;
  }
  let pred = null;
  let curr = root;
  while (curr) {
    if (node.val > curr.val) {
      pred = curr;
      curr = curr.right;
    } else {
      curr = curr.left;
    }
  }
  return pred;
}
    

This code finds the inorder predecessor of a given node in a BST by checking left subtree or ancestors.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing down the tree nodes using while loops.
• How many times: At most, it moves down the height of the tree twice: once to find the rightmost node in left subtree, or once to find the ancestor.

How Execution Grows With Input

As the tree grows taller, the number of steps to find the predecessor grows roughly with the tree height.

Input Size (n)	Approx. Operations
10	Up to about 3-4 steps (tree height ~3-4)
100	Up to about 7 steps (tree height ~7)
1000	Up to about 10 steps (tree height ~10)

Pattern observation: The steps grow slowly as the tree height grows, not directly with total nodes.

Final Time Complexity

Time Complexity: O(h)

This means the time depends on the height of the tree, which is how deep it is from root to leaf.

Common Mistake

[X] Wrong: "Finding the inorder predecessor always takes time proportional to the number of nodes (n)."

[OK] Correct: We only move down the tree height, not all nodes, so time depends on height (h), not total nodes (n).

Interview Connect

Understanding this helps you explain how tree shape affects search times, a key skill for working with BSTs and similar structures.

Self-Check

"What if the BST is balanced versus completely unbalanced? How would the time complexity change?"

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