BST Iterator Design in DSA Typescript

Mental Model

A BST iterator lets you see the tree's values one by one in order, like reading a book page by page without holding the whole book open.

Analogy: Imagine a bookmark that remembers where you stopped reading in a sorted book, so you can pick up exactly where you left off without rereading pages.

      7
     / \
    3   15
       /  \
      9    20

Stack: []
Iterator at start, ready to go leftmost.

Dry Run Walkthrough

Input: BST with nodes 7, 3, 15, 9, 20; iterate through all values in ascending order

Goal: Return values one by one in sorted order using next() and hasNext() methods

Step 1: Initialize iterator by pushing leftmost nodes onto stack

Stack: [7, 3]
Current node not yet returned

Why: We start at the smallest value by going left as far as possible

Step 2: Call next(), pop 3 from stack and return it

Stack: [7]
Returned value: 3

Why: 3 is the smallest node, so we return it first

Step 3: Call next(), pop 7 from stack and return it; push leftmost nodes of 7's right child (15)

Stack: [15, 9]
Returned value: 7

Why: After 7, we explore its right subtree starting at 15

Step 4: Call next(), pop 9 from stack and return it

Stack: [15]
Returned value: 9

Why: 9 is next smallest after 7

Step 5: Call next(), pop 15 from stack and return it; push leftmost nodes of 15's right child (20)

Stack: [20]
Returned value: 15

Why: After 15, move to its right child 20

Step 6: Call next(), pop 20 from stack and return it

Stack: []
Returned value: 20

Why: 20 is the largest node, iteration ends after this

Step 7: Call hasNext(), return false since stack is empty

Stack: []
No more nodes to return

Why: No nodes left means iteration is complete

Result:

Stack: []
Returned values in order: 3 -> 7 -> 9 -> 15 -> 20
Iteration complete

Annotated Code

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

class BSTIterator {
  private stack: TreeNode[] = [];

  constructor(root: TreeNode | null) {
    this.pushLeft(root);
  }

  private pushLeft(node: TreeNode | null): void {
    while (node !== null) {
      this.stack.push(node); // push current node to stack
      node = node.left; // move to left child
    }
  }

  next(): number {
    const node = this.stack.pop()!; // pop top node
    if (node.right !== null) {
      this.pushLeft(node.right); // push left path of right subtree
    }
    return node.val; // return current node's value
  }

  hasNext(): boolean {
    return this.stack.length > 0; // check if any nodes left
  }
}

// Driver code to test iterator
const root = new TreeNode(7,
  new TreeNode(3),
  new TreeNode(15, new TreeNode(9), new TreeNode(20))
);
const iterator = new BSTIterator(root);
const result: number[] = [];
while (iterator.hasNext()) {
  result.push(iterator.next());
}
console.log(result.join(' -> ') + ' -> null');

while (node !== null) { this.stack.push(node); // push current node to stack node = node.left; // move to left child }

push all left children to stack to reach smallest node

const node = this.stack.pop()!; // pop top node

pop the next smallest node to return

if (node.right !== null) { this.pushLeft(node.right); // push left path of right subtree }

after returning a node, prepare stack with left path of its right subtree

return node.val; // return current node's value

return the value of the current node

return this.stack.length > 0; // check if any nodes left

check if iteration can continue

Output

3 -> 7 -> 9 -> 15 -> 20 -> null

Complexity Analysis

Time: O(1) amortized per next() call because each node is pushed and popped once

Space: O(h) where h is tree height, due to stack storing left path nodes

vs Alternative: Compared to inorder traversal storing all nodes (O(n) space), this uses less memory by storing only path nodes

Edge Cases

Empty tree (root is null)

Iterator stack remains empty, hasNext() returns false immediately

DSA Typescript

while (node !== null) { this.stack.push(node); node = node.left; }

Tree with single node

Stack contains one node, next() returns that node's value, then hasNext() returns false

DSA Typescript

const node = this.stack.pop()!;

Tree with only left children

Stack initially contains all nodes, next() returns nodes in ascending order without pushing right children

DSA Typescript

while (node !== null) { this.stack.push(node); node = node.left; }

When to Use This Pattern

When asked to iterate over a BST in sorted order with next() and hasNext(), use a controlled stack to simulate inorder traversal lazily.

Common Mistakes

Mistake: Pushing all nodes of the tree onto stack at once, using O(n) space
Fix: Only push leftmost path nodes and push right subtree's left path when needed

Mistake: Not pushing left children of right subtree after popping a node
Fix: After popping a node, call pushLeft on its right child to maintain correct order

Summary

Provides an iterator to traverse BST nodes in ascending order using controlled stack.

Use when you want to access BST elements one by one without storing all nodes.

The key is to push leftmost nodes and update stack with right subtree's left path after each next().