Mirror a Binary Tree in DSA C# - Build from Scratch

Mirror a Binary Tree

📖 Scenario: Imagine you have a family tree represented as a binary tree. You want to create a mirror image of this tree, swapping left and right children at every node.

🎯 Goal: You will build a program that creates a binary tree, then mirrors it by swapping left and right children recursively, and finally prints the mirrored tree in order.

📋 What You'll Learn

Create a binary tree with specific nodes

Add a method to mirror the binary tree

Traverse and print the tree in order after mirroring

💡 Why This Matters

🌍 Real World

Mirroring trees is useful in graphics, image processing, and reversing hierarchical data structures.

💼 Career

Understanding tree manipulation and recursion is important for software engineers working with data structures, algorithms, and system design.

1

Create the Binary Tree Nodes

Create a class called Node with an integer Value and two Node references called Left and Right. Then create a binary tree with root node value 1, left child value 2, and right child value 3.

DSA C#

// Define the Node class and create the root node with children
// Your code here

Hint

Define a class with properties for value and left/right children. Then create the root and assign its left and right nodes.

2

Add Mirror Method

Add a method called Mirror that takes a Node parameter called node. This method should swap the Left and Right children of node and then recursively call itself on the children if they are not null.

DSA C#

public class Node
{
    public int Value;
    public Node Left;
    public Node Right;

    public Node(int value)
    {
        Value = value;
        Left = null;
        Right = null;
    }
}

Node root = new Node(1);
root.Left = new Node(2);
root.Right = new Node(3);

// Add the Mirror method below
// Your code here

Hint

Use a temporary variable to swap left and right children. Call Mirror recursively on children.

3

Add InOrder Traversal Method

Add a method called InOrder that takes a Node parameter called node. This method should print the Value of the nodes in in-order traversal (left, root, right).

DSA C#

public class Node
{
    public int Value;
    public Node Left;
    public Node Right;

    public Node(int value)
    {
        Value = value;
        Left = null;
        Right = null;
    }

    public static void Mirror(Node node)
    {
        if (node == null) return;

        Node temp = node.Left;
        node.Left = node.Right;
        node.Right = temp;

        Mirror(node.Left);
        Mirror(node.Right);
    }

    // Add InOrder method below
    // Your code here
}

Node root = new Node(1);
root.Left = new Node(2);
root.Right = new Node(3);

Hint

Use recursion to visit left child, print current node, then right child.

4

Mirror the Tree and Print InOrder

Call the Mirror method on the root node to mirror the tree. Then call the InOrder method on the root node to print the mirrored tree values in order.

DSA C#

public class Node
{
    public int Value;
    public Node Left;
    public Node Right;

    public Node(int value)
    {
        Value = value;
        Left = null;
        Right = null;
    }

    public static void Mirror(Node node)
    {
        if (node == null) return;

        Node temp = node.Left;
        node.Left = node.Right;
        node.Right = temp;

        Mirror(node.Left);
        Mirror(node.Right);
    }

    public static void InOrder(Node node)
    {
        if (node == null) return;

        InOrder(node.Left);
        System.Console.Write(node.Value + " ");
        InOrder(node.Right);
    }
}

Node root = new Node(1);
root.Left = new Node(2);
root.Right = new Node(3);

// Call Mirror and InOrder methods below
// Your code here

Hint

Call Node.Mirror(root); then Node.InOrder(root); to see the mirrored tree values.

Practice

1. What does mirroring a binary tree mean?

easy

A. Swapping the left and right children of every node

B. Removing all leaf nodes

C. Sorting the nodes in ascending order

D. Counting the total number of nodes

Solution

1. Step 1: Understand the mirror operation

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

2. Step 2: Identify the effect on the tree

   This operation flips the tree horizontally, creating a mirror image.

3. Final Answer:

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

4. Quick Check:

   Mirror = Swap children [OK]

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

Common Mistakes:

• Thinking mirror means sorting nodes
• Confusing mirror with deleting nodes
• Assuming mirror counts nodes

2. Which of the following C# method signatures correctly represents a recursive function to mirror a binary tree node named root?

easy

A. void MirrorTree(Node root)

B. int MirrorTree(Node root)

C. Node MirrorTree(Node root)

D. bool MirrorTree(Node root)

Solution

1. Step 1: Understand the purpose of the function

   The function modifies the tree in place and does not need to return a value.

2. Step 2: Choose the correct return type

   Since it changes the tree nodes directly, the return type should be void.

3. Final Answer:

   void MirrorTree(Node root) -> Option A

4. Quick Check:

   In-place change means void return [OK]

Hint: Mirror function usually returns void as it modifies tree directly [OK]

Common Mistakes:

• Returning int or bool unnecessarily
• Returning Node when not needed
• Using incorrect parameter types

3. Given the following binary tree and the C# code to mirror it, what will be the output of the inorder traversal after mirroring?

    1
   / \
  2   3
 /     \
4       5

void MirrorTree(Node root) {
  if (root == null) return;
  var temp = root.left;
  root.left = root.right;
  root.right = temp;
  MirrorTree(root.left);
  MirrorTree(root.right);
}

// Inorder traversal prints nodes in order: left, root, right

medium

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

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

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

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

Solution

1. Step 1: Mirror the tree structure

   Swap left and right children at every node:
   Original root 1: left=2, right=3 -> after mirror left=3, right=2
   Node 2: left=4, right=null -> after mirror left=null, right=4
   Node 3: left=null, right=5 -> after mirror left=5, right=null

2. Step 2: Perform inorder traversal on mirrored tree

   Inorder = left, root, right
   Start at root (1):
   Left subtree (3): left=5, right=null -> inorder: 5,3
   Root: 1
   Right subtree (2): left=null, right=4 -> inorder: 2,4
   Combined inorder: [5, 3, 1, 2, 4]

3. Final Answer:

   [5, 3, 1, 2, 4] -> Option C

4. Quick Check:

   Mirror + inorder = [5,3,1,2,4] [OK]

Hint: Mirror swaps children; inorder visits left-root-right [OK]

Common Mistakes:

• Not swapping children correctly
• Confusing preorder with inorder traversal
• Forgetting to mirror all nodes

4. The following C# code attempts to mirror a binary tree but does not work correctly. What is the main error?

void MirrorTree(Node root) {
  MirrorTree(root.left);
  MirrorTree(root.right);
  var temp = root.left;
  root.left = root.right;
  root.right = temp;
}

medium

A. Swapping children happens after recursive calls, causing incorrect mirror

B. Missing base case for null root

C. Using var instead of explicit type causes error

D. Function should return Node instead of void

Solution

1. Step 1: Analyze the base case

   The code lacks an if (root == null) return; check before recursing on children.

2. Step 2: Understand why this causes error

   Recursing on null children leads to NullReferenceException or stack overflow since there's no termination condition.

3. Final Answer:

   Missing base case for null root -> Option B

4. Quick Check:

   Recursion needs if(root == null) return; [OK]

Hint: Always include null base case in tree recursion [OK]

Common Mistakes:

• Omitting base case for null nodes
• Swapping after recursion instead of before
• Thinking var causes error

5. You have a binary tree where some nodes have only one child (either left or right). You want to mirror this tree using C#. Which approach correctly handles all cases including nodes with single children?

hard

A. Mirror by creating a new tree instead of modifying original

B. Only swap children if both left and right children exist

C. Swap children only at leaf nodes

D. Recursively swap left and right children if node is not null, then recurse on both children

Solution

1. Step 1: Consider nodes with single child

   Nodes with only one child still need swapping; swapping null with a child is valid.

2. Step 2: Choose recursive approach that swaps and recurses regardless of child presence

   Swapping children even if one is null ensures correct mirror. Then recurse on both children to continue mirroring.

3. Final Answer:

   Recursively swap left and right children if node is not null, then recurse on both children -> Option D

4. Quick Check:

   Swap children always, recurse both sides [OK]

Hint: Always swap children even if one is null, then recurse [OK]

Common Mistakes:

• Skipping swap when one child is null
• Swapping only leaf nodes
• Creating new tree unnecessarily