BST Find Maximum Element in DSA Go - Execution Trace

Concept Flow - BST Find Maximum Element

Start at root node

↓

Check if right child exists?

No→Current node is max

Yes↓

Move to right child

↩Back to check right child

Start at the root and keep moving to the right child until there is no right child left. The last node visited is the maximum.

Execution Sample

DSA Go

func FindMax(root *Node) *Node {
    current := root
    for current != nil && current.Right != nil {
        current = current.Right
    }
    return current
}

This code finds the maximum element in a BST by moving right until no right child exists.

Execution Table

Step	Operation	Current Node Value	Right Child Exists?	Pointer Changes	Visual State
1	Start at root	15	Yes	current = root (15)	15 / \ 10 20 / \ 17 25
2	Move right	20	Yes	current = current.Right (20)	15 / \ 10 20 / \ 17 25
3	Move right	25	No	current = current.Right (25)	15 / \ 10 20 / \ 17 25
4	Return current	25	No	Return node with value 25	Max element found: 25

💡 No right child at node 25, so 25 is the maximum element.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
current	root (15)	15	20	25	25

Key Moments - 3 Insights

Why do we keep moving to the right child to find the maximum?

In a BST, all values to the right are greater. The execution_table rows 1 to 3 show moving right until no right child exists, which means the current node is the largest.

What if the tree has only one node?

If the root has no right child (execution_table step 1 shows 'Right Child Exists?' as No), the root itself is the maximum.

Why do we stop when current.Right is nil, not when current is nil?

Stopping at current.Right == nil means current is the last node on the right path. If we moved to current == nil, we would lose the reference to the max node. See execution_table step 3 and 4.

Visual Quiz - 3 Questions

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

A 25

B 20

C 15

D 17

💡 Hint

Check the 'Current Node Value' column at step 2 in execution_table.

At which step does the algorithm stop moving right?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look for the step where 'Right Child Exists?' is No in execution_table.

If the root node had no right child, what would be the maximum element?

A The root node value

B The left child value

C Nil (no max)

D The right child value

💡 Hint

Refer to key_moments about single node tree and execution_table step 1.

Concept Snapshot

BST Find Maximum Element:
- Start at root node
- While right child exists, move right
- When no right child, current node is max
- Return current node
- Works because BST right subtree has larger values

Full Transcript

To find the maximum element in a Binary Search Tree (BST), start at the root node. Check if the current node has a right child. If yes, move to the right child and repeat. Continue this until the current node has no right child. This last node is the maximum element. This works because in a BST, all values in the right subtree are greater than the current node. The provided code uses a loop to move right until no right child exists, then returns the current node as the maximum. The execution table shows each step with the current node value and pointer changes. Key moments clarify why we move right and when to stop. The visual quiz tests understanding of the steps and conditions.