Create a Binary Tree Manually in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Manual Binary Tree Creation and Inorder Traversal

What is the output of the inorder traversal after manually creating this binary tree?

DSA C++

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

Node* root = new Node(10);
root->left = new Node(5);
root->right = new Node(15);
root->left->left = new Node(3);
root->left->right = new Node(7);

void inorder(Node* node) {
    if (!node) return;
    inorder(node->left);
    std::cout << node->data << " ";
    inorder(node->right);
}

inorder(root);

A 15 10 7 5 3

B 10 5 3 7 15

C 3 5 7 10 15

D 5 3 7 10 15

💡 Hint

Inorder traversal visits left subtree, then node, then right subtree.

Inorder traversal visits nodes in ascending order for a binary search tree. The tree nodes are 3, 5, 7, 10, 15 in that order.

🧠 Conceptual

intermediate

Number of Nodes in a Manually Created Binary Tree

If you manually create a binary tree with root node and add two children to the root, and then add one child to the left child, how many nodes does the tree have?

A 4

B 5

C 3

D 2

💡 Hint

Count the root, its two children, and the one child added to the left child.

The tree has 1 root + 2 children + 1 child of left child = 4 nodes total.

🔧 Debug

advanced

Identify the Bug in Manual Binary Tree Node Linking

What error will occur when running this code snippet that manually creates a binary tree?

DSA C++

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

Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);

// Mistake: Trying to access right child of a nullptr
std::cout << root->right->left->data << std::endl;

A No output, program runs fine

B Compilation error: missing semicolon

C Output: 0

D Segmentation fault (accessing nullptr)

💡 Hint

Check if root->right->left is initialized before accessing.

root->right->left is nullptr because no node was assigned. Accessing its data causes a segmentation fault.

🚀 Application

advanced

Manual Binary Tree Height Calculation

Given a manually created binary tree, what is the height of the tree?

DSA C++

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

Node* root = new Node(8);
root->left = new Node(3);
root->right = new Node(10);
root->left->left = new Node(1);
root->left->right = new Node(6);
root->left->right->left = new Node(4);
root->right->right = new Node(14);
root->right->right->left = new Node(13);

int height(Node* node) {
    if (!node) return 0;
    int leftHeight = height(node->left);
    int rightHeight = height(node->right);
    return 1 + (leftHeight > rightHeight ? leftHeight : rightHeight);
}

int h = height(root);
std::cout << h << std::endl;

A 3

B 4

C 5

D 2

💡 Hint

Height is the longest path from root to leaf node.

The longest path is root(8) -> right(10) -> right(14) -> left(13), which has 4 nodes, so height is 4.

🧠 Conceptual

expert

Number of Null Pointers in a Complete Binary Tree of Height 3

How many null pointers are there in a complete binary tree of height 3 (height counted as number of nodes on longest path from root to leaf)?

A 8

B 7

C 4

D 15

💡 Hint

A complete binary tree of height h has 2^h - 1 nodes. Number of null pointers = number of missing children.

A complete binary tree of height 3 has 7 nodes. Each node has 2 pointers, total 14 pointers. Since 7 nodes have 6 edges, null pointers = 14 - 6 = 8.