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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Mirroring a Simple Binary Tree

What is the output of the in-order traversal after mirroring the given binary tree?

DSA C++

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

void mirror(Node* root) {
    if (!root) return;
    mirror(root->left);
    mirror(root->right);
    Node* temp = root->left;
    root->left = root->right;
    root->right = temp;
}

void inorder(Node* root, std::vector<int>& result) {
    if (!root) return;
    inorder(root->left, result);
    result.push_back(root->data);
    inorder(root->right, result);
}

int main() {
    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);

    mirror(root);
    std::vector<int> result;
    inorder(root, result);
    for (int val : result) std::cout << val << " ";
    return 0;
}

A [5, 2, 4, 1, 3]

B [4, 2, 5, 1, 3]

C [3, 1, 4, 2, 5]

D [3, 1, 5, 2, 4]

💡 Hint

Remember that mirroring swaps left and right children at every node.

Mirroring swaps left and right children recursively. The original in-order traversal is [4, 2, 5, 1, 3]. After mirroring, the tree structure changes and the in-order traversal becomes [3, 1, 5, 2, 4].

🧠 Conceptual

intermediate

Understanding Mirror Operation on Binary Trees

Which statement correctly describes the effect of the mirror operation on a binary tree?

A It swaps the left and right children of every node in the tree.

B It reverses the values of nodes in the tree without changing structure.

C It removes all leaf nodes from the tree.

D It converts the binary tree into a linked list.

💡 Hint

Think about what happens to the children of each node during mirroring.

Mirroring a binary tree means swapping the left and right children of every node recursively, effectively creating a mirror image of the tree structure.

🔧 Debug

advanced

Identify the Bug in Mirror Function

What error will occur when running this mirror function code?

DSA C++

void mirror(Node* root) {
    if (root == nullptr) return;
    mirror(root->left);
    mirror(root->right);
    Node* temp = root->left;
    root->left = root->right;
    root->right = temp;
    root = nullptr;
}

A Memory leak due to not deleting nodes.

B Logical error: setting root to nullptr does not affect the tree structure.

C Runtime error: null pointer dereference.

D No error; function works correctly.

💡 Hint

Consider what setting root = nullptr inside the function does.

Setting root = nullptr only changes the local pointer variable and does not affect the actual tree nodes or structure. This line is unnecessary and misleading but does not cause a runtime error.

❓ Predict Output

advanced

Output After Mirroring a Complex Tree

Given the following tree and mirror function, what is the pre-order traversal output after mirroring?

DSA C++

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

void mirror(Node* root) {
    if (!root) return;
    mirror(root->left);
    mirror(root->right);
    Node* temp = root->left;
    root->left = root->right;
    root->right = temp;
}

void preorder(Node* root, std::vector<int>& result) {
    if (!root) return;
    result.push_back(root->data);
    preorder(root->left, result);
    preorder(root->right, result);
}

int main() {
    Node* root = new Node(10);
    root->left = new Node(20);
    root->right = new Node(30);
    root->left->left = new Node(40);
    root->left->right = new Node(50);
    root->right->right = new Node(60);

    mirror(root);
    std::vector<int> result;
    preorder(root, result);
    for (int val : result) std::cout << val << " ";
    return 0;
}

A [10, 20, 40, 50, 30, 60]

B [10, 30, 20, 60, 50, 40]

C [10, 30, 60, 20, 50, 40]

D [10, 20, 50, 40, 30, 60]

💡 Hint

Pre-order traversal visits root, then left subtree, then right subtree after mirroring.

After mirroring, left and right children are swapped at every node. The pre-order traversal visits nodes in the order: root, mirrored left subtree (original right), mirrored right subtree (original left).

🧠 Conceptual

expert

Effect of Mirroring on Tree Height and Structure

Which statement about the height and structure of a binary tree after mirroring is true?

A Mirroring does not change the height but reverses the left and right subtrees at every node.

B Mirroring changes the height of the tree but keeps the node values in the same order.

C Mirroring flattens the tree into a linked list increasing the height.

D Mirroring duplicates all nodes, doubling the tree size.

💡 Hint

Think about what swapping left and right children does to the tree shape and height.

Mirroring swaps left and right children at every node, which does not change the height of the tree but reverses the structure horizontally.