Validate if Tree is BST in DSA Javascript - Execution Trace

Concept Flow - Validate if Tree is BST

Start at root node

↓

Check node value within allowed range?

Yes↓

Recurse left subtree with updated max bound

↓

Recurse right subtree with updated min bound

↓

If all nodes valid, return True

↓

Else return False

Start from the root, check if each node's value fits the BST rules within allowed min and max bounds, recursively check left and right subtrees.

Execution Sample

DSA Javascript

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 are within valid min and max bounds recursively.

Execution Table

Step	Operation	Node Visited	Allowed Range (min, max)	Result	Tree State
1	Start at root	10	(-∞, ∞)	Check 10 in range	10 / \ 5 15
2	Recurse left	5	(-∞, 10)	Check 5 in range	10 / \ 5 15
3	Recurse left.left	3	(-∞, 5)	Check 3 in range	10 / \ 5 15 / 3
4	Recurse left.left.left	null	(-∞, 3)	Null node, return true	No change
5	Recurse left.left.right	null	(3, 5)	Null node, return true	No change
6	Back to 5, recurse right	7	(5, 10)	Check 7 in range	10 / \ 5 15 / \ 3 7
7	Recurse right.left	null	(5, 7)	Null node, return true	No change
8	Recurse right.right	null	(7, 10)	Null node, return true	No change
9	Back to root, recurse right	15	(10, ∞)	Check 15 in range	10 / \ 5 15 / \ 3 7
10	Recurse right.left	13	(10, 15)	Check 13 in range	10 / \ 5 15 / \ / 3 7 13
11	Recurse right.left.left	null	(10, 13)	Null node, return true	No change
12	Recurse right.left.right	null	(13, 15)	Null node, return true	No change
13	Recurse right.right	20	(15, ∞)	Check 20 in range	10 / \ 5 15 / \ / \ 3 7 13 20
14	Recurse right.right.left	null	(15, 20)	Null node, return true	No change
15	Recurse right.right.right	null	(20, ∞)	Null node, return true	No change
16	All nodes valid	N/A	N/A	Return true	Tree is BST

💡 All nodes satisfy BST property within their allowed ranges, so the tree is a valid BST.

Variable Tracker

Variable	Start	After Step 2	After Step 6	After Step 9	After Step 13	Final
node.val	10	5	7	15	20	N/A
min	-∞	-∞	5	10	15	N/A
max	∞	10	10	∞	∞	N/A
Return value	N/A	true	true	true	true	true

Key Moments - 3 Insights

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

Because in a BST, all left subtree nodes must be less than the current node, so max is current node's value; all right subtree nodes must be greater, so min is current node's value. See steps 2 and 9 in execution_table.

What happens when we reach a null node?

A null node means leaf's child, which is valid by default, so we return true. See steps 4, 5, 7, 8, 11, 12, 14, 15.

Why do we check node.val <= min or node.val >= max instead of just equality?

Because BST requires strictly less or greater values, duplicates are not allowed on either side. This ensures strict ordering. See step 3 where node.val is checked against min and max.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what is the allowed range for node 7?

A (5, 10)

B (-∞, 10)

C (5, 7)

D (7, 10)

💡 Hint

Check the 'Allowed Range' column at step 6 in execution_table.

At which step does the recursion check the right child of node 15?

A Step 10

B Step 9

C Step 13

D Step 16

💡 Hint

Look for 'Recurse right.right' operation in execution_table.

If node 13 had value 16, which step would fail the BST check?

A Step 10

B Step 13

C Step 6

D Step 2

💡 Hint

Check the allowed range for node 13 at step 10 and what happens at step 13.

Concept Snapshot

Validate BST by checking each node's value is strictly between min and max bounds.
Start at root with (-∞, ∞).
Recurse left with max = node.val.
Recurse right with min = node.val.
Return false if any node violates bounds.
Return true if all nodes valid.

Full Transcript

To check if a binary tree is a BST, we start at the root node and verify if its value lies within an allowed range, initially from negative infinity to positive infinity. For each node, we recursively check the left subtree with an updated maximum bound equal to the current node's value, and the right subtree with an updated minimum bound equal to the current node's value. If any node's value is not strictly within these bounds, the tree is not a BST. Null nodes are considered valid leaves. This process continues until all nodes are validated or a violation is found. The example tree with nodes 10, 5, 15, 3, 7, 13, and 20 passes all checks and is a valid BST.