Mirror a Binary Tree in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

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

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

Achild

Bleft

Cright

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using root.left instead of root.right causes no swap.

Trying to swap with a non-existent property.

Swapping root.left with root.right requires accessing root.right to assign it to root.left.

2fill in blank

medium

Complete the code to recursively call mirror on the left subtree.

DSA Java

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

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

Achild

Bnode

Cright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Calling mirror on root.right instead of root.left only mirrors half the tree.

Calling mirror on a non-existent property.

To mirror the left subtree, call mirror on root.left.

3fill in blank

hard

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

DSA Java

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

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

Aright

Bchild

Cleft

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Calling mirror on root.left twice.

Using an incorrect property name.

To mirror the right subtree, call mirror on root.right.

4fill in blank

hard

Fill both blanks to complete the swap and recursive calls for mirroring.

DSA Java

TreeNode temp = root.[1];
root.[2] = root.right;
root.right = temp;
mirror(root.left);
mirror(root.right);

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

Aleft

Bright

Cchild

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the wrong child nodes.

Not swapping before recursive calls.

Swap root.left with root.right by storing root.left in temp and assigning root.right to root.left.

5fill in blank

hard

Fill all three blanks to complete the mirror function with base case and recursive calls.

DSA Java

public void mirror(TreeNode [1]) {
    if ([2] == null) return;
    TreeNode temp = [2].left;
    [2].left = [2].right;
    [2].right = temp;
    mirror([2].left);
    mirror([2].right);
}

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

Anode

Broot

CtreeNode

DrootNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causing errors.

Missing the base case check.

The parameter and all references inside the function use the same variable name 'root' for clarity.

Practice

1. What does it mean to mirror a binary tree?

easy

A. Remove all leaf nodes from the tree

B. Swap the left and right children of every node in the tree

C. Convert the tree into a linked list

D. Sort the nodes in ascending order

Solution

1. Step 1: Understand the mirror operation

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

2. Step 2: Apply the definition to the tree

   By swapping children at every node, the tree becomes a mirror image of the original.

3. Final Answer:

   Swap the left and right children of every node in the tree -> Option B

4. Quick Check:

   Mirror = Swap children [OK]

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

Common Mistakes:

• Thinking mirror means deleting nodes
• Confusing mirror with sorting
• Assuming mirror flattens the tree

2. Which of the following Java method signatures correctly represents a function to mirror a binary tree node?

easy

A. public void mirrorTree(Node root)

B. public int mirrorTree(Node root)

C. void mirrorTree(Node root) throws Exception

D. public Node mirrorTree()

Solution

1. Step 1: Identify correct return type

   Mirroring modifies the tree in place, so void return type is appropriate.

2. Step 2: Check method parameters

   The method should accept the root node to start mirroring from.

3. Final Answer:

   public void mirrorTree(Node root) -> Option A

4. Quick Check:

   Void return + root parameter = correct signature [OK]

Hint: Mirror modifies tree in place, so use void with root parameter [OK]

Common Mistakes:

• Using return type int instead of void
• Missing root parameter
• Adding unnecessary exceptions

3. Given the following binary tree and the mirror function below, what will be the preorder traversal output after mirroring?

    1
   / \
  2   3
 /     \
4       5

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

medium

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

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

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

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

Solution

1. Step 1: Understand the mirror function

   The function swaps left and right children at every node recursively.

2. Step 2: Apply mirror to the tree

   Original tree preorder: 1,2,4,3,5. After mirror, left and right children swap: - Root 1: left=3, right=2 - Node 3: left=5, right=null - Node 2: left=null, right=4

3. Step 3: Preorder traversal after mirror

   Visit root, then left subtree, then right subtree: 1, 3, 5, 2, 4

4. Final Answer:

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

5. Quick Check:

   Mirror swaps children; preorder = [1,3,5,2,4] [OK]

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

Common Mistakes:

• Not swapping children correctly
• Confusing preorder with inorder
• Missing recursive calls

4. The following code attempts to mirror a binary tree but causes a NullPointerException. What is the error?

public void mirrorTree(Node root) {
  Node temp = root.left;
  root.left = root.right;
  root.right = temp;
  mirrorTree(root.left);
  mirrorTree(root.right);
}

medium

A. Swapping left and right children incorrectly

B. Recursive calls should be after swapping

C. Missing null check before accessing root.left

D. Method should return a Node instead of void

Solution

1. Step 1: Identify cause of NullPointerException

   The code accesses root.left without checking if root is null.

2. Step 2: Fix by adding null check

   Adding "if (root == null) return;" at the start prevents errors when root is null.

3. Final Answer:

   Missing null check before accessing root.left -> Option C

4. Quick Check:

   NullPointerException = missing null check [OK]

Hint: Always check if node is null before accessing children [OK]

Common Mistakes:

• Ignoring null checks
• Thinking swap order causes error
• Changing return type unnecessarily

5. You have a binary tree where some nodes have only one child. After mirroring the tree using the standard mirror function, which of the following statements is true about the structure of the mirrored tree?

hard

A. Leaf nodes become internal nodes after mirroring

B. Nodes with one child will have two children after mirroring

C. The tree height will always increase by 1

D. Nodes with only left child will have only right child after mirroring

Solution

1. Step 1: Understand mirroring effect on children

   Mirroring swaps left and right children at every node.

2. Step 2: Analyze nodes with one child

   If a node has only a left child, after mirroring it will have only a right child (the left child moves to right).

3. Step 3: Check other options

   Tree height does not necessarily increase; leaf nodes remain leaves; nodes do not gain extra children.

4. Final Answer:

   Nodes with only left child will have only right child after mirroring -> Option D

5. Quick Check:

   Mirror swaps children, so left-only becomes right-only [OK]

Hint: Mirror swaps children, so single left child becomes single right child [OK]

Common Mistakes:

• Assuming tree height changes
• Thinking nodes gain extra children
• Confusing leaf and internal nodes