Validate if Tree is BST in DSA C++ - Execution Trace

Concept Flow - Validate if Tree is BST

Start at root node

↓

Check left subtree

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Is left subtree BST?

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Check current node value

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Check right subtree

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Is right subtree BST?

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Return True

We check recursively if left subtree is BST, then current node value fits BST rules, then right subtree is BST.

Execution Sample

DSA C++

bool isBST(Node* node, int min, int max) {
  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 code checks if a binary tree is a BST by verifying node values lie within valid min and max ranges.

Execution Table

Step	Operation	Node Visited	Min Allowed	Max Allowed	Check Result	Return Value	Tree State
1	Start at root	10	-∞	+∞	10 > -∞ and 10 < +∞	Pending	[10]
2	Check left subtree	5	-∞	10	5 > -∞ and 5 < 10	Pending	[10, 5]
3	Check left subtree of 5	2	-∞	5	2 > -∞ and 2 < 5	Pending	[10, 5, 2]
4	Check left subtree of 2	null	-∞	2	null node	True	[10, 5, 2]
5	Check right subtree of 2	null	2	5	null node	True	[10, 5, 2]
6	Return from node 2	2	-∞	5	Both subtrees True	True	[10, 5, 2]
7	Check right subtree of 5	7	5	10	7 > 5 and 7 < 10	Pending	[10, 5, 2, 7]
8	Check left subtree of 7	null	5	7	null node	True	[10, 5, 2, 7]
9	Check right subtree of 7	null	7	10	null node	True	[10, 5, 2, 7]
10	Return from node 7	7	5	10	Both subtrees True	True	[10, 5, 2, 7]
11	Return from node 5	5	-∞	10	Both subtrees True	True	[10, 5, 2, 7]
12	Check right subtree	15	10	+∞	15 > 10 and 15 < +∞	Pending	[10, 5, 2, 7, 15]
13	Check left subtree of 15	12	10	15	12 > 10 and 12 < 15	Pending	[10, 5, 2, 7, 15, 12]
14	Check left subtree of 12	null	10	12	null node	True	[10, 5, 2, 7, 15, 12]
15	Check right subtree of 12	null	12	15	null node	True	[10, 5, 2, 7, 15, 12]
16	Return from node 12	12	10	15	Both subtrees True	True	[10, 5, 2, 7, 15, 12]
17	Check right subtree of 15	20	15	+∞	20 > 15 and 20 < +∞	Pending	[10, 5, 2, 7, 15, 12, 20]
18	Check left subtree of 20	null	15	20	null node	True	[10, 5, 2, 7, 15, 12, 20]
19	Check right subtree of 20	null	20	+∞	null node	True	[10, 5, 2, 7, 15, 12, 20]
20	Return from node 20	20	15	+∞	Both subtrees True	True	[10, 5, 2, 7, 15, 12, 20]
21	Return from node 15	15	10	+∞	Both subtrees True	True	[10, 5, 2, 7, 15, 12, 20]
22	Return from root 10	10	-∞	+∞	Both subtrees True	True	[10, 5, 2, 7, 15, 12, 20]

💡 All nodes satisfy BST conditions, recursion ends with True

Variable Tracker

Variable	Start	After Step 2	After Step 6	After Step 11	After Step 21	Final
node	10 (root)	5 (left child)	5 (left child)	10 (root)	15 (right child)	10 (root)
min	-∞	-∞	-∞	-∞	10	-∞
max	+∞	10	10	10	+∞	+∞
return value	Pending	Pending	True	True	True	True

Key Moments - 3 Insights

Why do we pass min and max values when checking each node?

The min and max track the allowed value range for each node. See execution_table steps 2, 12 where min and max change to keep BST rules.

What happens when we reach a null node?

Null nodes mean leaf ends and are considered valid BST parts. See steps 4, 5, 8 where null nodes return True.

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

BST requires strict less or greater, no duplicates allowed. See step 3 where node value must be strictly between min and max.

Visual Quiz - 3 Questions

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

A min = 5, max = 10

B min = -∞, max = 10

C min = 2, max = 7

D min = 7, max = 10

💡 Hint

Check columns 'Min Allowed' and 'Max Allowed' at step 7 in execution_table

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

A Step 6

B Step 11

C Step 21

D Step 22

💡 Hint

Look for 'Return from node 5' with return value True in execution_table

If node 12 had value 9 instead, which step would fail the BST check?

A Step 16

B Step 12

C Step 13

D Step 17

💡 Hint

Check step 13 where node 12 is checked against min=10 and max=15

Concept Snapshot

Validate BST by recursion:
- Start at root with min=-∞, max=+∞
- Check node value in (min, max)
- Recurse left with max=node.val
- Recurse right with min=node.val
- Null nodes return True
- If any check fails, return False

Full Transcript

To check if a binary tree is a BST, we start at the root node and verify if its value lies within allowed minimum and maximum bounds. Initially, these bounds are negative and positive infinity. We then recursively check the left subtree, updating the maximum bound to the current node's value, and the right subtree, updating the minimum bound to the current node's value. If a node is null, it is considered valid. If any node violates the bounds, the tree is not a BST. The execution table shows each step visiting nodes, checking values against bounds, and returning True or False accordingly. The variable tracker follows the current node and bounds during recursion. Key moments clarify why bounds are needed, how null nodes are handled, and why strict inequalities are used. The visual quiz tests understanding of bounds at specific steps and failure points. This method ensures the entire tree satisfies BST properties.