Tree Traversal Preorder Root Left Right in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Preorder Traversal on a Simple Binary Tree

What is the output of the preorder traversal (Root, Left, Right) of the following binary tree?

Tree structure:
1
/ \
2 3
/ \
4 5

DSA C++

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

void preorder(Node* root) {
    if (!root) return;
    std::cout << root->val << " ";
    preorder(root->left);
    preorder(root->right);
}

int main() {
    Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->right->left = new Node(4);
    root->right->right = new Node(5);
    preorder(root);
    return 0;
}

A 1 2 4 3 5

B 1 3 4 5 2

C 1 2 3 4 5

D 2 1 4 3 5

💡 Hint

Remember preorder visits root first, then left subtree, then right subtree.

Preorder traversal visits nodes in the order: root, left subtree, right subtree. Starting at 1, then left child 2, then right subtree 3, 4, 5.

🧠 Conceptual

intermediate

Number of Nodes Visited in Preorder Traversal

Given a binary tree with 7 nodes, how many nodes will be visited during a preorder traversal?

A 8

B 6

C Depends on the tree structure

D 7

💡 Hint

Preorder traversal visits every node exactly once.

Preorder traversal visits every node in the tree exactly once, so all 7 nodes will be visited.

❓ Predict Output

advanced

Output of Preorder Traversal with Null Children

What is the output of the preorder traversal of this tree?

Tree structure:
10
/ \
null 20
/ \
30 null

DSA C++

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

void preorder(Node* root) {
    if (!root) return;
    std::cout << root->val << " ";
    preorder(root->left);
    preorder(root->right);
}

int main() {
    Node* root = new Node(10);
    root->right = new Node(20);
    root->right->left = new Node(30);
    preorder(root);
    return 0;
}

A 10 20 30

B 10 30 20

C 20 10 30

D 10 20

💡 Hint

Preorder visits root, then left, then right. Null children are skipped.

Starting at 10, left child is null so skipped, then right child 20, then 20's left child 30.

🔧 Debug

advanced

Identify the Error in Preorder Traversal Code

What error will this preorder traversal code produce when run on a non-empty tree?

DSA C++

void preorder(Node* root) {
    if (root == nullptr) return;
    preorder(root->left);
    std::cout << root->val << " ";
    preorder(root->right);
}

A Compilation error due to missing return type

B Prints nodes in inorder, not preorder

C Prints nodes in preorder correctly

D Runtime error due to null pointer dereference

💡 Hint

Check the order of printing and recursive calls.

The code prints left subtree first, then root, then right subtree, which is inorder traversal, not preorder.

🧠 Conceptual

expert

Preorder Traversal Output for a Skewed Tree

Consider a binary tree where every node has only a right child, forming a chain of nodes with values 1 to 5. What is the preorder traversal output?

A 2 1 3 4 5

B 1 3 5 2 4

C 1 2 3 4 5

D 5 4 3 2 1

💡 Hint

Preorder visits root first, then left (none), then right.

Since each node has only a right child, preorder visits nodes in order 1 to 5.