BST Find Maximum Element in DSA C++

Mental Model

In a Binary Search Tree, the biggest number is always the rightmost node. We just keep going right until we can't go further.

Analogy: Imagine a line of boxes arranged from smallest on the left to biggest on the right. To find the biggest box, you just walk to the rightmost box in the line.

      10
     /  \
    5    15
          \
           20
            \
             25

Start at root (10) and keep moving right to find max.

Dry Run Walkthrough

Input: BST with nodes: 10 -> 5 (left), 15 (right), 15 -> 20 (right), 20 -> 25 (right)

Goal: Find the maximum element in the BST

Step 1: Start at root node 10

10 [curr->10] -> 5 (left), 15 (right)

Why: We begin at the root to find the maximum

Step 2: Move curr to right child 15

10 -> 15 [curr->15] -> 20 (right)

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

Step 3: Move curr to right child 20

10 -> 15 -> 20 [curr->20] -> 25 (right)

Why: Keep moving right to find bigger values

Step 4: Move curr to right child 25

10 -> 15 -> 20 -> 25 [curr->25] -> null

Why: 25 has no right child, so it is the maximum

Result:

10 -> 15 -> 20 -> 25 [curr->25] -> null
Maximum element is 25

Annotated Code

DSA C++

#include <iostream>
#include <stdexcept>
using namespace std;

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

int findMax(Node* root) {
    if (root == nullptr) {
        throw runtime_error("Tree is empty");
    }
    Node* curr = root;
    while (curr->right != nullptr) {
        curr = curr->right; // advance curr to right child to find max
    }
    return curr->data; // curr is rightmost node, max value
}

int main() {
    Node* root = new Node(10);
    root->left = new Node(5);
    root->right = new Node(15);
    root->right->right = new Node(20);
    root->right->right->right = new Node(25);

    cout << "Maximum element is " << findMax(root) << endl;
    return 0;
}

while (curr->right != nullptr) {

keep moving curr to right child to find the maximum

curr = curr->right; // advance curr to right child to find max

advance curr pointer to next right node

return curr->data; // curr is rightmost node, max value

return the data of the rightmost node as maximum

Output

Maximum element is 25

Complexity Analysis

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

Space: O(1) because we use only a few pointers and no extra space proportional to input

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

Edge Cases

Empty tree (root is nullptr)

Throws an error indicating the tree is empty

DSA C++

if (root == nullptr) { throw runtime_error("Tree is empty"); }

Tree with only one node

Returns that single node's value as maximum

DSA C++

while (curr->right != nullptr) { ... } // loop skipped if no right child

When to Use This Pattern

When you see a BST and need the largest value, look for the rightmost node by following right children until none remain.

Common Mistakes

Mistake: Trying to find max by checking all nodes instead of just going right
Fix: Use the BST property and move only to the right child until null

Mistake: Not handling empty tree and causing null pointer errors
Fix: Add a check for root == nullptr and handle it properly

Summary

Finds the maximum element by moving to the rightmost node in the BST.

Use when you want the largest value in a BST quickly.

The maximum is always the rightmost node because BST nodes to the right are bigger.