Mirror a Binary Tree in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap the left and right child nodes of the root.

DSA C#

var temp = root.left;
root.left = root.[1];
root.right = temp;

Drag options to blanks, or click blank then click option'

Achild

Bleft

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with root.left again instead of root.right.

Using incorrect property names like parent or child.

Swapping root.left with root.right mirrors the root's immediate children.

2fill in blank

medium

Complete the code to recursively mirror the left subtree.

DSA C#

if (root.left != null) {
    [1](root.left);
}

Drag options to blanks, or click blank then click option'

APrintTree

BTraverseTree

CInvertTree

DMirrorTree

Attempts:

3 left

💡 Hint

Common Mistakes

Calling a wrong function like PrintTree or TraverseTree.

Not calling recursively at all.

The function MirrorTree is called recursively on the left child to mirror the subtree.

3fill in blank

hard

Fix the error in the recursive call to mirror the right subtree.

DSA C#

if (root.right != null) {
    [1](root.right);
}

Drag options to blanks, or click blank then click option'

AMirrorTree

BMirrorTrees

CMirror

DMirrorTreeRight

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect function names like MirrorTrees or MirrorTreeRight.

Misspelling the function name.

The correct function name is MirrorTree to recursively mirror the right subtree.

4fill in blank

hard

Fill both blanks to complete the mirror function that swaps children and recursively mirrors subtrees.

DSA C#

public void MirrorTree(Node root) {
    if (root == null) return;
    var temp = root.[1];
    root.[2] = root.left;
    root.left = temp;
    if (root.left != null) MirrorTree(root.left);
    if (root.right != null) MirrorTree(root.right);
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with wrong properties like parent or child.

Mixing up left and right assignments.

First, store root.right in temp, then assign root.right to root.left, and finally assign temp to root.left to swap children.

5fill in blank

hard

Fill all three blanks to complete the mirror function that swaps children and recursively mirrors both subtrees.

DSA C#

public void MirrorTree(Node root) {
    if (root == null) return;
    var temp = root.[1];
    root.[2] = root.left;
    root.left = temp;
    if (root.left != null) [3](root.left);
    if (root.right != null) MirrorTree(root.right);
}

Drag options to blanks, or click blank then click option'

Aright

BMirrorTree

Cleft

DInvertTree

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong function names for recursion.

Swapping with wrong child properties.

Swap root.right with root.left, then recursively call MirrorTree on the left child to mirror subtrees.

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