BST Iterator Design in DSA Typescript - Execution Trace

Concept Flow - BST Iterator Design

Initialize stack empty

↓

Push all left nodes from root

↓

Check if stack empty?

Yes→No next element

No↓

Pop top node from stack

↓

Return popped node's value

↓

Push all left nodes from popped node's right child

↓

Repeat check for next element

The iterator uses a stack to store nodes. It pushes all left children first, then pops nodes to return values, pushing left children of right subtrees as it goes.

Execution Sample

DSA Typescript

class BSTIterator {
  stack: TreeNode[] = [];
  constructor(root: TreeNode) {
    this.pushLeft(root);
  }
  pushLeft(node: TreeNode | null) {
    while (node) {
      this.stack.push(node);
      node = node.left;
    }
  }
  next(): number {
    const node = this.stack.pop()!;
    this.pushLeft(node.right);
    return node.val;
  }
  hasNext(): boolean {
    return this.stack.length > 0;
  }
}

This code creates an iterator over a BST that returns nodes in ascending order using a stack to track left nodes.

Execution Table

Step	Operation	Stack Nodes (top to bottom)	Pointer Changes	Visual State
1	Initialize iterator with root=7	[1, 3, 7]	Push 7, then 3, then 1	Stack: 1(top) -> 3 -> 7 -> null
2	Call hasNext()	Stack unchanged	No pointer change	Stack: 1(top) -> 3 -> 7 -> null
3	Call next()	Pop 1	Pop 1; push left nodes of 1.right (null)	Returned 1; Stack: 3(top) -> 7 -> null
4	Call hasNext()	Stack unchanged	No pointer change	Stack: 3(top) -> 7 -> null
5	Call next()	Pop 3	Pop 3; push left nodes of 3.right (5)	Returned 3; Stack: 5(top) -> 7 -> null
6	Call hasNext()	Stack unchanged	No pointer change	Stack: 5(top) -> 7 -> null
7	Call next()	Pop 5	Pop 5; push left nodes of 5.right (null)	Returned 5; Stack: 7(top) -> null
8	Call hasNext()	Stack unchanged	No pointer change	Stack: 7(top) -> null
9	Call next()	Pop 7	Pop 7; push left nodes of 7.right (9)	Returned 7; Stack: 9(top) -> null
10	Call hasNext()	Stack unchanged	No pointer change	Stack: 9(top) -> null
11	Call next()	Pop 9	Pop 9; push left nodes of 9.right (null)	Returned 9; Stack: empty
12	Call hasNext()	Stack empty	No pointer change	Stack: empty
13	Call next()	No nodes to pop	No pointer change	No next element, iteration ends

💡 Stack becomes empty after all nodes are visited; hasNext() returns false and next() cannot proceed.

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 5	After Step 7	After Step 9	After Step 11	Final
stack	[]	[7,3,1]	[7,3]	[7,5]	[7]	[9]	[]	[]
current node popped	null	null	1	3	5	7	9	null
returned value	null	null	1	3	5	7	9	null

Key Moments - 3 Insights

Why do we push all left children onto the stack at initialization and after popping a node?

Because the smallest values are always in the left subtree, pushing all left children ensures the next smallest node is on top of the stack, as shown in steps 1 and 9 in the execution_table.

What happens if the popped node has no right child?

No new nodes are pushed onto the stack after popping, so the stack size decreases, as seen in steps 3 and 7 where no nodes are pushed after popping nodes 1 and 5.

Why does hasNext() check if the stack is empty?

Because the stack holds all nodes yet to be visited; if empty, no next node exists. This is clear in step 12 where the stack is empty and hasNext() returns false.

Visual Quiz - 3 Questions

Look at the execution_table at step 5. What is the top node on the stack after the next() call?

A Node with value 5

B Node with value 3

C Node with value 7

D Stack is empty

💡 Hint

Refer to the 'Stack Nodes' column at step 5 in the execution_table.

At which step does the stack become empty, indicating no more elements to iterate?

A Step 11

B Step 12

C Step 13

D Step 10

💡 Hint

Check the 'Stack Nodes' column and 'exit_note' in the execution_table.

If the BST had an additional left child under node 1, how would the stack change after initialization (step 1)?

A Stack would include the new left child on top

B Stack would remain the same

C Stack would only have the root node

D Stack would be empty

💡 Hint

Refer to the 'pushLeft' operation in the code and how left children are pushed onto the stack.

Concept Snapshot

BST Iterator uses a stack to traverse nodes in ascending order.
Initialize by pushing all left children from root.
next() pops top node, pushes left children of its right subtree.
hasNext() checks if stack is not empty.
This simulates in-order traversal without recursion.

Full Transcript

The BST Iterator design uses a stack to simulate in-order traversal of a binary search tree. Initially, it pushes all left children starting from the root onto the stack. Each call to next() pops the top node from the stack, returns its value, and then pushes all left children of the popped node's right child. The hasNext() method returns true if the stack is not empty, indicating more nodes to visit. This approach allows iteration over the BST in ascending order without recursion, maintaining the current traversal state in the stack.