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DSA Goprogramming~5 mins

BST Find Maximum Element in DSA Go - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is a Binary Search Tree (BST)?
A Binary Search Tree is a tree data structure where each node has at most two children. The left child's value is less than the parent's value, and the right child's value is greater than the parent's value.
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beginner
In a BST, where is the maximum element located?
The maximum element in a BST is located at the rightmost node because values increase as you move right.
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beginner
Why do we keep moving right to find the maximum element in a BST?
Because in a BST, all right children have values greater than their parents, so moving right leads to larger values until the largest is found.
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intermediate
What is the time complexity of finding the maximum element in a BST?
The time complexity is O(h), where h is the height of the BST. In the worst case (skewed tree), it can be O(n), and in a balanced tree, it is O(log n).
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intermediate
Show the Go code snippet to find the maximum element in a BST.
func findMax(root *Node) *Node { if root == nil { return nil } current := root for current.Right != nil { current = current.Right } return current }
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Where do you find the maximum element in a BST?
AAny leaf node
BRightmost node
CRoot node
DLeftmost node
What is the first step to find the maximum element in a BST?
AGo to the right child
BGo to the left child
CCheck the root value
DTraverse all nodes
If a BST has only one node, what is the maximum element?
ANo maximum element
BLeft child
CThe root node itself
DRight child
What is the time complexity to find the maximum element in a balanced BST with n nodes?
AO(n)
BO(1)
CO(n log n)
DO(log n)
What happens if you call findMax on an empty BST?
AReturns nil
BReturns root
CReturns error
DReturns zero
Explain how to find the maximum element in a Binary Search Tree.
Think about the BST property where right children are greater.
You got /4 concepts.
    Write a simple Go function to find the maximum element in a BST and explain its steps.
    Use a loop to move right until no more right child.
    You got /4 concepts.