Mirror a Binary Tree in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap the left and right children of the root node.

DSA Javascript

function mirrorTree(root) {
  if (!root) return null;
  const temp = root.left;
  root.left = root.[1];
  root.right = temp;
  return root;
}

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

Aright

Bleft

Cchild

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 Javascript

function mirrorTree(root) {
  if (!root) return null;
  root.left = mirrorTree(root.[1]);
  root.right = mirrorTree(root.right);
  const temp = root.left;
  root.left = root.right;
  root.right = temp;
  return root;
}

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

Aright

Bchild

Cparent

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Recursively calling mirrorTree on root.right instead of root.left.

Using incorrect property names.

We recursively call mirrorTree on root.left to mirror the left subtree before swapping.

3fill in blank

hard

Fix the error in the base case to correctly handle an empty tree.

DSA Javascript

function mirrorTree(root) {
  if (root === [1]) return null;
  root.left = mirrorTree(root.left);
  root.right = mirrorTree(root.right);
  const temp = root.left;
  root.left = root.right;
  root.right = temp;
  return root;
}

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

Anull

Bfalse

Cundefined

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for undefined instead of null.

Using 0 or false which are not valid tree node checks.

The base case should check if root is null to stop recursion.

4fill in blank

hard

Fill both blanks to complete the recursive mirror function correctly.

DSA Javascript

function mirrorTree(root) {
  if (root === [1]) return null;
  root.left = mirrorTree(root.[2]);
  root.right = mirrorTree(root.right);
  const temp = root.left;
  root.left = root.right;
  root.right = temp;
  return root;
}

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

Anull

Bleft

Cright

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined instead of null for base case.

Recursively calling mirrorTree on root.right instead of root.left.

Base case checks for null; recursively mirror left subtree using root.left.

5fill in blank

hard

Fill all three blanks to implement a complete mirror function with recursion and swapping.

DSA Javascript

function mirrorTree(root) {
  if (root === [1]) return null;
  root.left = mirrorTree(root.[2]);
  root.right = mirrorTree(root.[3]);
  const temp = root.left;
  root.left = root.right;
  root.right = temp;
  return root;
}

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

Anull

Bleft

Cright

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Missing base case or using undefined.

Not recursively calling mirrorTree on both children.

Not swapping children after recursion.

Check for null base case; recursively mirror left and right subtrees; then swap children.

Practice

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

easy

A. Sort the nodes in ascending order

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

C. Convert the tree into a linked list

D. Remove all leaf nodes from the tree

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 swapping 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 is the correct JavaScript syntax to swap the left and right children of a node named node?

easy

A. let temp = node.left; node.left = node.right; node.right = temp;

B. node.left = node.right; node.right = node.left;

C. node.left, node.right = node.right, node.left;

D. swap(node.left, node.right);

Solution

1. Step 1: Understand swapping variables in JavaScript

   To swap two variables, use a temporary variable to hold one value during the swap.

2. Step 2: Check each option

   let temp = node.left; node.left = node.right; node.right = temp; correctly uses a temporary variable to swap node.left and node.right. node.left = node.right; node.right = node.left; overwrites node.left before assigning node.right. node.left, node.right = node.right, node.left; uses invalid syntax in JavaScript. swap(node.left, node.right); calls an undefined function.

3. Final Answer:

   let temp = node.left; node.left = node.right; node.right = temp; -> Option A

4. Quick Check:

   Use temp variable to swap [OK]

Hint: Use a temp variable to swap two values in JavaScript [OK]

Common Mistakes:

• Overwriting one child before saving it
• Using Python-style tuple swap syntax
• Calling undefined swap functions

3. What is the output of the following code that mirrors a binary tree?

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function mirror(node) {
  if (!node) return null;
  let left = mirror(node.left);
  let right = mirror(node.right);
  node.left = right;
  node.right = left;
  return node;
}

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

let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.right.right = new Node(5);

mirror(root);

function printTree(node) {
  if (!node) return 'null';
  return node.val + ' -> ' + printTree(node.left) + ', ' + printTree(node.right);
}

console.log(printTree(root));

medium

A. 1 -> 2 -> 4 -> null, 3 -> 5 -> null, null

B. 1 -> 3 -> 4 -> null, 2 -> 5 -> null, null

C. 1 -> 3 -> null, 2 -> 5 -> null, 4

D. 1 -> 3 -> 5 -> null, 2 -> 4 -> null, null

Solution

1. Step 1: Understand the mirror function

   The mirror function swaps left and right children recursively for every node.

2. Step 2: Trace the tree after mirroring

   Original tree: 1 with left=2 and right=3; 2 has left=4; 3 has right=5.
   After mirroring:
   1's left becomes 3, right becomes 2.
   3's left becomes 5, right becomes null.
   2's left becomes null, right becomes 4.
   Printing in preorder with left then right shows: 1 -> 3 -> 5 -> null, 2 -> 4 -> null, null.

3. Final Answer:

   1 -> 3 -> 5 -> null, 2 -> 4 -> null, null -> Option D

4. Quick Check:

   Mirror swaps children recursively [OK]

Hint: Trace swaps at each node recursively [OK]

Common Mistakes:

• Not swapping children at all nodes
• Mixing left and right children incorrectly
• Misreading the print output format

4. Find the error in the following mirror function and select the correct fix:

function mirror(node) {
  if (node == null) return;
  mirror(node.left);
  mirror(node.right);
  let temp = node.left;
  node.left = node.right;
  node.right = temp;
}

medium

A. The function should return the node after swapping children

B. The base case should check for undefined instead of null

C. The swap should happen before recursive calls

D. The temp variable should be declared outside the function

Solution

1. Step 1: Analyze the function behavior

   The function swaps children after recursive calls but does not return the node, so the caller gets undefined.

2. Step 2: Identify the fix

   Returning the node after swapping ensures the mirrored subtree is returned properly for further use.

3. Final Answer:

   The function should return the node after swapping children -> Option A

4. Quick Check:

   Return node after swap [OK]

Hint: Always return the node after modifying it in recursion [OK]

Common Mistakes:

• Not returning the node causes undefined results
• Swapping before recursion breaks logic
• Misunderstanding null vs undefined checks

5. Given a binary tree where some nodes have only one child, how does the mirror function handle these nodes?
Consider the tree:

    10
   /  
  20   
    \
     30

What will be the structure after mirroring?

hard

A. 10 with left child 30 and right child 20; 20 and 30 have no children

B. 10 with left child 20 and right null; 20 with left null and right 30

C. 10 with left child null and right child 20; 20 with left child 30 and right null

D. 10 with left child null and right child 30; 30 with left child 20 and right null

Solution

1. Step 1: Understand mirroring nodes with single children

   Mirroring swaps left and right children even if one child is null.

2. Step 2: Apply mirror to given tree

   Original:
   10 has left=20, right=null
   20 has left=null, right=30
   After mirror:
   10's left becomes null, right becomes 20
   20's left becomes 30, right becomes null

3. Final Answer:

   10 with left child null and right child 20; 20 with left child 30 and right null -> Option C

4. Quick Check:

   Mirror swaps children even if one is null [OK]

Hint: Swap children even if one child is missing [OK]

Common Mistakes:

• Ignoring null children during swap
• Assuming children move to wrong side
• Confusing node values with positions