BST Find Maximum Element in DSA Go

Mental Model

In a BST, the biggest number is always the rightmost node. Just keep going right until you can't go further.

Analogy: Imagine a line of people standing from shortest to tallest, left to right. The tallest person is at the far right end.

      10
     /  \
    5    15
          \
           20
            \
             25
↑root

Dry Run Walkthrough

Input: BST with nodes 10 -> 5, 15 -> 20 -> 25; find max element

Goal: Find the largest value in the BST by moving right until no more right child

Step 1: Start at root node 10

10 -> [curr->15] -> 20 -> 25
5 (left child of 10)

Why: We begin at the root to find the maximum

Step 2: Move curr pointer from 10 to right child 15

10 -> 15 -> [curr->20] -> 25
5 (left child of 10)

Why: Maximum is always in the right subtree, so move right

Step 3: Move curr pointer from 15 to right child 20

10 -> 15 -> 20 -> [curr->25]
5 (left child of 10)

Why: Keep moving right to find bigger values

Step 4: Move curr pointer from 20 to right child 25

10 -> 15 -> 20 -> 25 [curr]
5 (left child of 10)

Why: Continue moving right until no more right child

Step 5: No right child of 25, stop and return 25

10 -> 15 -> 20 -> 25 [curr]
5 (left child of 10)

Why: Rightmost node found, this is the maximum

Result:

10 -> 15 -> 20 -> 25 [max]
5 (left child of 10)
Maximum element is 25

Annotated Code

DSA Go

package main

import "fmt"

// Node represents a BST node
type Node struct {
    val   int
    left  *Node
    right *Node
}

// Insert inserts a value into the BST
func (n *Node) Insert(val int) *Node {
    if n == nil {
        return &Node{val: val}
    }
    if val < n.val {
        n.left = n.left.Insert(val)
    } else {
        n.right = n.right.Insert(val)
    }
    return n
}

// FindMax finds the maximum value in the BST
func (n *Node) FindMax() int {
    curr := n
    for curr.right != nil {
        curr = curr.right // move right to find bigger values
    }
    return curr.val // rightmost node value is max
}

func main() {
    var root *Node
    values := []int{10, 5, 15, 20, 25}
    for _, v := range values {
        root = root.Insert(v)
    }
    maxVal := root.FindMax()
    fmt.Printf("Maximum element is %d\n", maxVal)
}

curr := n

start at root node

for curr.right != nil {

keep moving right while right child exists

curr = curr.right // move right to find bigger values

advance curr to right child to find max

return curr.val // rightmost node value is max

return value of rightmost node

Output

Maximum element is 25

Complexity Analysis

Time: O(h) because we only traverse down the right children where h is the tree height

Space: O(1) because we use only a pointer variable and no extra space

vs Alternative: Compared to scanning all nodes (O(n)), this is faster as it uses BST property to go directly right

Edge Cases

Empty tree (root is nil)

Function is not called or handled outside; no max to find

DSA Go

No explicit guard in FindMax; caller must check root != nil

Tree with single node

FindMax returns that node's value immediately

DSA Go

for curr.right != nil { ... } loop exits immediately

All nodes only on left side (no right children)

FindMax returns root value as no right child exists

DSA Go

for curr.right != nil { ... } loop exits immediately

When to Use This Pattern

When asked to find the largest element in a BST, look for the rightmost node by following right children until none remain.

Common Mistakes

Mistake: Trying to search both left and right subtrees for max instead of just going right
Fix: Use BST property and only move right until no right child

Mistake: Not handling empty tree and calling FindMax on nil
Fix: Check if root is nil before calling FindMax

Summary

Finds the largest value in a BST by moving right until no more right child.

Use when you need the maximum element quickly in a BST.

The maximum is always the rightmost node in a BST.