BST Inorder Successor in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the left child of a node.

DSA C++

TreeNode* leftChild = node->[1];

Drag options to blanks, or click blank then click option'

Aparent

Bleft

Cright

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' will give the right child, not the left.

Using 'parent' or 'next' is incorrect as they are not standard BST node pointers.

The left child of a node in a BST is accessed using node->left.

2fill in blank

medium

Complete the code to find the minimum node in a BST subtree.

DSA C++

while (current->[1] != nullptr) {
    current = current->[1];
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Checking right child instead of left will find the maximum, not minimum.

Using parent or next pointers is incorrect for this traversal.

To find the minimum node in a BST subtree, keep moving to the left child until it is null.

3fill in blank

hard

Fix the error in the code to find the inorder successor when the node has no right child.

DSA C++

TreeNode* successor = nullptr;
TreeNode* ancestor = root;
while (ancestor != node) {
    if (node->val < ancestor->val) {
        successor = ancestor;
        ancestor = ancestor->[1];
    } else {
        ancestor = ancestor->right;
    }
}

Drag options to blanks, or click blank then click option'

Aparent

Bright

Cnext

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Moving to right child instead of left will cause infinite loop or wrong successor.

Using parent or next pointers is invalid here without parent links.

When the node has no right child, move up the tree via ancestors. If node's value is less than ancestor's, go left.

4fill in blank

hard

Fill both blanks to complete the function that finds the inorder successor in a BST.

DSA C++

if (node->right != nullptr) {
    TreeNode* temp = node->right;
    while (temp->[1] != nullptr) {
        temp = temp->[2];
    }
    return temp;
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' will move to the right subtree, not the leftmost node.

Using 'parent' or 'next' is incorrect without parent pointers.

The inorder successor in the right subtree is the leftmost node. So keep moving left.

5fill in blank

hard

Fill all three blanks to complete the full inorder successor function in a BST without parent pointers.

DSA C++

TreeNode* inorderSuccessor(TreeNode* root, TreeNode* node) {
    if (node->right != nullptr) {
        TreeNode* temp = node->right;
        while (temp->[1] != nullptr) {
            temp = temp->[2];
        }
        return temp;
    }
    TreeNode* successor = nullptr;
    TreeNode* ancestor = root;
    while (ancestor != node) {
        if (node->val < ancestor->val) {
            successor = ancestor;
            ancestor = ancestor->[3];
        } else {
            ancestor = ancestor->right;
        }
    }
    return successor;
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cparent

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using right pointers instead of left to find minimum node.

Using parent or next pointers which are not available.

Incorrect traversal direction when searching for successor.

First, find the leftmost node in the right subtree. If no right child, traverse from root to find the lowest ancestor whose left child is also ancestor of node.