BST Find Minimum Element in DSA Javascript - Execution Trace

Concept Flow - BST Find Minimum Element

Start at root node

↓

Check if left child exists?

No→Current node is minimum

Yes↓

Move to left child

↩Back to check left child

Start at the root and keep moving left until no left child exists; that node is the minimum.

Execution Sample

DSA Javascript

function findMin(root) {
  let current = root;
  while (current.left !== null) {
    current = current.left;
  }
  return current.value;
}

This code finds the smallest value in a BST by moving left until no left child remains.

Execution Table

Step	Operation	Current Node Value	Left Child Exists?	Action	Visual State
1	Start at root	10	Yes	Move to left child	10 / \ 5 15
2	At node 5	5	Yes	Move to left child	10 / \ 5 15 / 2
3	At node 2	2	No	Stop, found minimum	10 / \ 5 15 / 2
Exit	Minimum found	2	N/A	Return 2	Final minimum node is 2

💡 Left child does not exist at node 2, so 2 is the minimum.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
current	root (10)	5	2	2	2

Key Moments - 3 Insights

Why do we keep moving left to find the minimum?

Because in a BST, smaller values are always on the left. The execution_table shows moving left from 10 to 5, then 5 to 2, until no left child exists.

What if the tree has only one node?

The loop never runs because current.left is null at the start. The execution_table step 1 would show no left child, so the root is the minimum.

Why do we stop when current.left is null, not when current is null?

Stopping at current.left null means current is the smallest node. If we went further, current would become null and lose the node value. See step 3 in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'current' after step 2?

A 2

B 5

C 10

D 15

💡 Hint

Check the 'current' variable in variable_tracker after step 2.

At which step does the algorithm stop moving left?

A Step 1

B Step 3

C Step 2

D Exit

💡 Hint

Look at the 'Left Child Exists?' column in execution_table.

If the root node had no left child, what would the minimum be?

A Null

B The right child's value

C The root node's value

D The smallest right child's leftmost node

💡 Hint

Refer to key_moments about single-node or no-left-child trees.

Concept Snapshot

BST Find Minimum Element:
- Start at root node
- Move left while left child exists
- When no left child, current node is minimum
- Return current node's value
- Works because BST left children are smaller

Full Transcript

To find the minimum element in a Binary Search Tree (BST), start at the root node. Check if the current node has a left child. If yes, move to that left child and repeat. Continue moving left until you reach a node with no left child. That node holds the minimum value in the BST. This works because BSTs store smaller values on the left side. The code uses a loop to move left until current.left is null, then returns current.value. If the tree has only one node or the root has no left child, the root itself is the minimum. Stopping when current.left is null ensures we do not lose the node's value by moving beyond the smallest node.