BST Find Minimum Element in DSA C++ - Execution Trace

Concept Flow - BST Find Minimum Element

Start at root

↓

Check if current.left exists?

No→Current node is minimum

Yes↓

Move current to current.left

↩Back to check

Start at the root node and keep moving left until no left child exists. The last node reached is the minimum.

Execution Sample

DSA C++

Node* findMin(Node* root) {
  Node* current = root;
  while (current != nullptr && current->left != nullptr) {
    current = current->left;
  }
  return current;
}

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

Execution Table

Step	Operation	Current Node	Pointer Changes	Visual State
1	Start at root	50	current = root (50)	50 / \ 30 70 / \ \ 20 40 80
2	Check if current.left exists	50	Yes, move current to 30	50 / \ 30 70 / \ \ 20 40 80
3	Check if current.left exists	30	Yes, move current to 20	50 / \ 30 70 / \ \ 20 40 80
4	Check if current.left exists	20	No left child, stop	50 / \ 30 70 / \ \ 20 40 80
5	Return current as minimum	20	Return node with value 20	Minimum element found: 20

💡 Stopped when current node (20) has no left child

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
current	50 (root)	30	20	20	20

Key Moments - 3 Insights

Why do we keep moving left to find the minimum?

In a BST, smaller values are always in the left subtree. So moving left leads to the smallest value. See execution_table steps 2 and 3 where current moves left.

What if the tree is empty (root is nullptr)?

The code checks if current is nullptr before accessing left. If root is nullptr, current is nullptr and the loop doesn't run, returning nullptr immediately.

Why do we stop when current->left is nullptr, not when current is nullptr?

Because current itself is a valid node holding the minimum value. If current were nullptr, it means we went past the leaf. See execution_table step 4 where current is 20 and has no left child.

Visual Quiz - 3 Questions

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

A 20

B 50

C 30

D 40

💡 Hint

Check the 'current' column in execution_table row for step 3.

At which step does the condition 'current->left != nullptr' become false?

A Step 3

B Step 4

C Step 2

D Step 5

💡 Hint

Look at the 'Operation' and 'Pointer Changes' columns in execution_table for step 4.

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

A nullptr

B The right child of root

C The root node itself

D The leftmost leaf

💡 Hint

Refer to the concept_flow and execution_table step 1 and 4 where no left child means current node is minimum.

Concept Snapshot

BST Find Minimum Element:
- Start at root node
- Move left child repeatedly
- Stop when no left child
- Current node is minimum
- Returns pointer to minimum node

Full Transcript

To find the minimum element in a Binary Search Tree (BST), start at the root node. Because smaller values are always on the left, keep moving to the left child node until you reach a node that has no left child. That node holds the minimum value. The code uses a pointer 'current' starting at root and moves left while current->left is not null. When current->left is null, current points to the minimum node. If the tree is empty, the function returns nullptr immediately. This process ensures the smallest element is found efficiently.