BST Find Minimum Element in DSA Go - Execution Trace

Concept Flow - BST Find Minimum Element

Start at root

↓

Check if left child exists?

No→Current node is minimum

Yes↓

Move to left child

↓

Repeat check

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

Execution Sample

DSA Go

func findMin(root *Node) *Node {
    current := root
    for current.Left != nil {
        current = current.Left
    }
    return current
}

This code finds the minimum element in a BST by moving left until no left child is found.

Execution Table

Step	Operation	Current Node	Left Child Exists?	Pointer Changes	Visual State
1	Start at root	20	Yes	current = root (20)	20 / \ 10 30
2	Check left child	20	Yes	Move current to left child (10)	20 / \ 10 30
3	Check left child	10	Yes	Move current to left child (5)	20 / \ 10 30 / 5
4	Check left child	5	No	Stop, current is minimum	20 / \ 10 30 / 5
5	Return current	5	No	Return node with value 5	Minimum element found: 5

💡 Left child does not exist at node 5, so 5 is the minimum element.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
current	20 (root)	10	5	5	5

Key Moments - 3 Insights

Why do we keep moving to the left child to find the minimum?

Because in a BST, all smaller values are on the left. The execution_table rows 2 and 3 show moving left to smaller nodes until no left child exists.

What if the root has no left child?

Then the root itself is the minimum. The execution_table row 1 shows the root node with a left child, but if it had none, we would stop immediately as in row 4.

Why do we stop when current.Left is nil?

Because no smaller node exists to the left. The execution_table row 4 shows this stopping condition.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'current' after step 3?

A 5

B 10

C 20

D 30

💡 Hint

Check the 'Pointer Changes' and 'Current Node' columns at step 3.

At which step does the condition 'Left Child Exists?' become false?

A Step 3

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'Left Child Exists?' column to find when it changes to 'No'.

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

A The left child of root

B The right child of root

C The root node itself

D No minimum element

💡 Hint

Refer to the key moment about root having no left child and the execution_table row 4.

Concept Snapshot

BST Find Minimum Element:
- Start at root node
- While left child exists, move left
- When no left child, current node is minimum
- Return current node
- Works because BST left subtree has smaller values

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. Repeat this process until you reach a node with no left child. This node is the minimum element in the BST. The code uses a loop to move left until current.Left is nil, then returns the current node. This works because in BSTs, smaller values are always to the left. If the root has no left child, the root itself is the minimum. The execution table shows each step with the current node and pointer changes, confirming the minimum found.