Validate if Tree is BST in DSA Typescript - Execution Trace

Concept Flow - Validate if Tree is BST

Start at root node

↓

Check if node is null?

Yes→Return True

No↓

Check node value > min and < max?

No→Return False

Yes↓

Recurse left subtree with updated max = node value

↓

Recurse right subtree with updated min = node value

↓

Return left_result AND right_result

We start at the root and check if each node's value fits within allowed min and max limits, updating these limits as we go down left and right subtrees.

Execution Sample

DSA Typescript

function isBST(node, min = -Infinity, max = Infinity) {
  if (!node) return true;
  if (node.val <= min || node.val >= max) return false;
  return isBST(node.left, min, node.val) && isBST(node.right, node.val, max);
}

This function checks if a binary tree is a BST by verifying node values against min and max bounds recursively.

Execution Table

Step	Operation	Node Visited	Min Bound	Max Bound	Result	Tree State
1	Start at root	10	-Infinity	Infinity	Continue	10 / \ 5 15
2	Check node 10 value	10	-Infinity	Infinity	10 > -∞ and < ∞: True	10 / \ 5 15
3	Recurse left subtree	5	-Infinity	10	Continue	10 / \ 5 15
4	Check node 5 value	5	-Infinity	10	5 > -∞ and < 10: True	10 / \ 5 15
5	Recurse left subtree	2	-Infinity	5	Continue	10 / \ 5 15 / 2
6	Check node 2 value	2	-Infinity	5	2 > -∞ and < 5: True	10 / \ 5 15 / 2
7	Recurse left subtree	null	-Infinity	2	True (null node)	10 / \ 5 15 / 2
8	Recurse right subtree	null	2	5	True (null node)	10 / \ 5 15 / 2
9	Left subtree of 5 is BST	5	-Infinity	10	True	10 / \ 5 15 / 2
10	Recurse right subtree	7	5	10	Continue	10 / \ 5 15 / \ 2 7
11	Check node 7 value	7	5	10	7 > 5 and < 10: True	10 / \ 5 15 / \ 2 7
12	Recurse left subtree	null	5	7	True (null node)	10 / \ 5 15 / \ 2 7
13	Recurse right subtree	null	7	10	True (null node)	10 / \ 5 15 / \ 2 7
14	Right subtree of 5 is BST	5	-Infinity	10	True	10 / \ 5 15 / \ 2 7
15	Left subtree of 10 is BST	10	-Infinity	Infinity	True	10 / \ 5 15 / \ 2 7
16	Recurse right subtree	15	10	Infinity	Continue	10 / \ 5 15 / \ 2 7 20
17	Check node 15 value	15	10	Infinity	15 > 10 and < ∞: True	10 / \ 5 15 / \ 2 7 20
18	Recurse left subtree	null	10	15	True (null node)	10 / \ 5 15 / \ 2 7 20
19	Recurse right subtree	20	15	Infinity	Continue	10 / \ 5 15 / \ 2 7 20
20	Check node 20 value	20	15	Infinity	20 > 15 and < ∞: True	10 / \ 5 15 / \ 2 7 20
21	Recurse left subtree	null	15	20	True (null node)	10 / \ 5 15 / \ 2 7 20
22	Recurse right subtree	null	20	Infinity	True (null node)	10 / \ 5 15 / \ 2 7 20
23	Right subtree of 15 is BST	15	10	Infinity	True	10 / \ 5 15 / \ 2 7 20
24	Right subtree of 10 is BST	10	-Infinity	Infinity	True	10 / \ 5 15 / \ 2 7 20
25	Final result	root	-Infinity	Infinity	True (Tree is BST)	10 / \ 5 15 / \ 2 7 20

💡 All nodes satisfy BST conditions; recursion ends with True.

Variable Tracker

Variable	Start	After Step 3	After Step 9	After Step 15	After Step 24	Final
node	10 (root)	5 (left child)	5 (left child)	10 (root)	15 (right child)	10 (root)
min	-Infinity	-Infinity	-Infinity	-Infinity	10	-Infinity
max	Infinity	10	10	Infinity	Infinity	Infinity
result	N/A	Continue	True (left subtree BST)	True (left subtree BST)	True (right subtree BST)	True (whole tree BST)

Key Moments - 3 Insights

Why do we update max when going left and min when going right?

Because left subtree nodes must be less than the current node (max updated to node.val), and right subtree nodes must be greater (min updated to node.val). See execution_table steps 3 and 16.

What happens when we reach a null node?

We return True because an empty subtree is always a valid BST. This is shown in steps 7, 8, 12, 13, 18, 21, and 22.

Why do we check node.val <= min or >= max to return False?

Because if node value is not strictly between min and max, it violates BST rules. This check ensures no duplicates or invalid placements. See step 2 for the condition.

Visual Quiz - 3 Questions

Look at the execution_table at step 10, what is the min and max bounds for node 7?

A min = 7, max = 10

B min = 5, max = 10

C min = -Infinity, max = 5

D min = 10, max = Infinity

💡 Hint

Check the 'Min Bound' and 'Max Bound' columns at step 10 in execution_table.

At which step does the recursion confirm the left subtree of node 5 is a BST?

A Step 9

B Step 14

C Step 15

D Step 24

💡 Hint

Look for the row where 'Left subtree of 5 is BST' is mentioned in the 'Operation' column.

If node 7 had value 11, what would happen in the execution_table?

A Step 10 min and max bounds would change

B Step 11 would continue as normal

C Step 11 would return False because 11 is not < 10

D The tree would still be a BST

💡 Hint

Check the condition at step 11 where node value must be between min and max.

Concept Snapshot

Validate if Tree is BST:
- Start at root with min = -∞, max = ∞
- For each node, check if min < node.val < max
- Recurse left with max = node.val
- Recurse right with min = node.val
- Return True if all nodes satisfy conditions
- Return False immediately if violation found

Full Transcript

To check if a binary tree is a BST, we start at the root node with minimum and maximum bounds set to negative and positive infinity. For each node, we verify if its value is strictly greater than the min bound and less than the max bound. We then recursively check the left subtree, updating the max bound to the current node's value, and the right subtree, updating the min bound to the current node's value. If any node violates these bounds, we return false immediately. If we reach a null node, we return true since an empty subtree is valid. The process continues until all nodes are validated, confirming if the tree is a BST or not.