Why Mirror a Binary Tree in DSA C++?

The Big Idea

What if you could flip a whole tree upside down with just a few lines of code?

The Scenario

Imagine you have a family tree drawn on paper, and you want to see its reflection as if looking in a mirror. Doing this by hand means redrawing every branch and leaf on the opposite side, which is tiring and easy to mess up.

The Problem

Manually swapping every left and right branch is slow and error-prone. You might forget some branches or place them incorrectly, making the mirrored tree wrong and confusing.

The Solution

Mirroring a binary tree with code swaps the left and right children of every node automatically. This simple process repeats down the tree, giving a perfect mirror image quickly and without mistakes.

Before vs After

✗ Before

void mirrorTree(Node* root) {
  // Manually swap left and right for each node
  if (root == nullptr) return;
  Node* temp = root->left;
  root->left = root->right;
  root->right = temp;
  mirrorTree(root->left);
  mirrorTree(root->right);
}

✓ After

void mirrorTree(Node* root) {
  if (!root) return;
  std::swap(root->left, root->right);
  mirrorTree(root->left);
  mirrorTree(root->right);
}

What It Enables

This lets you quickly create a mirrored version of any tree, useful in graphics, data transformations, and problem solving.

Real Life Example

In photo editing apps, flipping an image horizontally is like mirroring a tree structure of pixels, making the picture look reversed instantly.

Key Takeaways

Manual mirroring is slow and error-prone.

Code swaps left and right children recursively for a perfect mirror.

Useful in many computer graphics and data tasks.

Practice

1. What does mirroring a binary tree mean?

easy

A. Sorting the nodes in ascending order

B. Reversing the order of nodes in each level

C. Swapping the left and right children of every node

D. Deleting all leaf nodes

Solution

1. Step 1: Understand the mirror operation

   Mirroring means swapping left and right children of every node in the tree.

2. Step 2: Clarify what happens to nodes

   Each node's left child becomes its right child and vice versa, recursively.

3. Final Answer:

   Swapping the left and right children of every node -> Option C

4. Quick Check:

   Mirror = Swap children [OK]

Hint: Mirror means swap left and right children at all nodes [OK]

Common Mistakes:

• Thinking mirror means reversing node values
• Confusing mirror with tree traversal order
• Assuming mirror deletes nodes

2. Which of the following is the correct C++ function signature to mirror a binary tree node?

easy

A. void mirror(Node* root)

B. int mirror(Node root)

C. Node mirror(Node* root)

D. void mirror(Node root)

Solution

1. Step 1: Identify correct parameter type

   The function should take a pointer to the root node to modify the tree in place.

2. Step 2: Identify correct return type

   Since the tree is modified in place, the function can return void.

3. Final Answer:

   void mirror(Node* root) -> Option A

4. Quick Check:

   Pointer param + void return = void mirror(Node* root) [OK]

Hint: Use pointer parameter and void return for in-place tree changes [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Returning Node instead of void
• Using incorrect parameter types

3. What is the output of the following code after mirroring the binary tree?

struct Node {
  int data;
  Node* left;
  Node* right;
};

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

// Tree before mirror:
//     1
//    / \
//   2   3
//  /     \
// 4       5

Node* root = new Node{1, new Node{2, new Node{4, nullptr, nullptr}, nullptr}, new Node{3, nullptr, new Node{5, nullptr, nullptr}}};
mirror(root);
// Print inorder traversal
void inorder(Node* node) {
  if (!node) return;
  inorder(node->left);
  std::cout << node->data << " ";
  inorder(node->right);
}
inorder(root);

medium

A. 5 3 1 4 2

B. 5 3 1 2 4

C. 1 2 4 3 5

D. 4 2 1 3 5

Solution

1. Step 1: Understand the original tree and inorder traversal

   Original inorder: left-root-right = 4 2 1 3 5

2. Step 2: Mirror swaps left and right children recursively

   Mirrored tree structure: 1 / \ 3 2 / \ 5 4 Inorder traversal now: 5 3 1 2 4

3. Final Answer:

   5 3 1 2 4 -> Option B

4. Quick Check:

   Inorder mirrored = 5 3 1 2 4 [OK]

Hint: Mirror swaps children; inorder reverses left-right order [OK]

Common Mistakes:

• Not swapping children correctly
• Confusing preorder with inorder traversal
• Printing original tree instead of mirrored

4. Identify the error in this mirror function and choose the fix:

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;
  return root;
}

medium

A. Swap root->left and root->right before recursive calls

B. Change 'if (root == nullptr)' to 'if (!root->left && !root->right)'

C. Add 'delete temp;' after swapping

D. Function should return void, remove 'return root;'

Solution

1. Step 1: Check function return type and usage

   The function is declared void but tries to return root pointer, which causes a compile error.

2. Step 2: Fix by removing the return statement

   Since the tree is modified in place, no return is needed; remove 'return root;'.

3. Final Answer:

   Function should return void, remove 'return root;' -> Option D

4. Quick Check:

   Void function must not return value [OK]

Hint: Void functions should not return values [OK]

Common Mistakes:

• Returning value from void function
• Changing base case incorrectly
• Deleting nodes unnecessarily

5. Given a binary tree, how would you modify the mirror function to also count how many nodes were swapped during the mirroring process?

hard

A. Return the number of nodes from the function and sum left and right counts

B. Add a counter parameter and increment it each time children are swapped

C. Use a global variable to count swaps inside the mirror function

D. Count nodes before and after mirroring and subtract counts

Solution

1. Step 1: Understand counting swaps during recursion

   Each node swap can be counted by returning 1 plus counts from left and right subtrees.

2. Step 2: Modify mirror to return int and sum counts

   Return 0 if node is null; else return 1 + mirror(left) + mirror(right).

3. Final Answer:

   Return the number of nodes from the function and sum left and right counts -> Option A

4. Quick Check:

   Return counts from recursion to sum swaps [OK]

Hint: Return counts from recursion to sum swapped nodes [OK]

Common Mistakes:

• Using global variables causing side effects
• Counting nodes before and after mirroring (no change)
• Not accumulating counts correctly