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#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);

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);
}

List<int> result = new List<int>();
void InOrder(Node node) {
    if (node == null) return;
    InOrder(node.left);
    result.Add(node.val);
    InOrder(node.right);
}

Mirror(root);
InOrder(root);
Console.WriteLine(string.Join(",", result));

A 3,2,1

B 3,1,2

C 2,1,3

D 1,3,2

💡 Hint

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

Mirroring swaps left and right children. Original in-order: 2,1,3. After mirror, left and right children of root swap, so in-order becomes 3,1,2.

❓ Predict Output

intermediate

Mirror a Binary Tree with Multiple Levels

What is the pre-order traversal output after mirroring this binary tree?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

Node root = new Node(4);
root.left = new Node(2);
root.right = new Node(7);
root.left.left = new Node(1);
root.left.right = new Node(3);
root.right.left = new Node(6);
root.right.right = new Node(9);

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);
}

List<int> result = new List<int>();
void PreOrder(Node node) {
    if (node == null) return;
    result.Add(node.val);
    PreOrder(node.left);
    PreOrder(node.right);
}

Mirror(root);
PreOrder(root);
Console.WriteLine(string.Join(",", result));

A 4,7,9,6,2,3,1

B 4,2,1,3,7,6,9

C 4,7,6,9,2,1,3

D 4,9,7,6,3,2,1

💡 Hint

Pre-order visits root, then left, then right. Mirroring swaps left and right children.

After mirroring, left and right children swap at every node. The pre-order traversal visits root first, then left subtree, then right subtree. The mirrored tree's pre-order is 4,7,9,6,2,3,1.

🔧 Debug

advanced

Identify the Error in Mirror Function

What error will this mirror function cause when run on a binary tree?

DSA C#

void Mirror(Node node) {
    if (node == null) return;
    Node temp = node.left;
    node.left = node.right;
    node.right = temp;
    Mirror(node.right);
    Mirror(node.left);
}

A No error; function works correctly

B StackOverflow due to infinite recursion

C NullReferenceException at runtime

D Compilation error due to wrong syntax

💡 Hint

Check the order of recursive calls after swapping children.

The function swaps left and right children, then recursively calls Mirror on right and left children (which were swapped). This still covers all nodes once, so no infinite recursion or errors occur.

❓ Predict Output

advanced

Output of Level-Order Traversal After Mirroring

Given the binary tree below, what is the output of the level-order traversal after mirroring?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

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);
root.right.left = new Node(12);
root.right.right = new Node(18);

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);
}

List<int> result = new List<int>();
Queue<Node> queue = new Queue<Node>();

Mirror(root);
queue.Enqueue(root);
while(queue.Count > 0) {
    Node current = queue.Dequeue();
    result.Add(current.val);
    if (current.left != null) queue.Enqueue(current.left);
    if (current.right != null) queue.Enqueue(current.right);
}

Console.WriteLine(string.Join(",", result));

A 10,5,15,3,7,12,18

B 10,5,15,7,3,18,12

C 10,15,5,12,18,3,7

D 10,15,5,18,12,7,3

💡 Hint

Level-order visits nodes level by level from left to right. Mirroring swaps children at every node.

After mirroring, left and right children swap. The level order traversal visits nodes level by level, so the output is 10,15,5,18,12,7,3.

🧠 Conceptual

expert

Effect of Mirroring Twice on a Binary Tree

What is the result of applying the mirror operation twice on any binary tree?

A The tree becomes a binary search tree

B The tree becomes a complete binary tree

C The tree remains unchanged, identical to the original

D The tree becomes a linked list

💡 Hint

Think about what happens when you swap left and right children twice.

Mirroring swaps left and right children. Doing it twice swaps them back to original positions, so the tree remains unchanged.

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