Why Mirror a Binary Tree in DSA Javascript?

The Big Idea

What if you could flip a whole tree upside down with just a few lines of code?

The Scenario

Imagine you have a family tree drawn on paper, and you want to see its reflection as if looking in a mirror. Doing this by hand means redrawing every branch and swapping left and right children manually.

The Problem

Manually swapping each left and right child is slow and easy to mess up, especially for big trees. You might forget a node or swap incorrectly, causing confusion and errors.

The Solution

Mirroring a binary tree with code automatically swaps left and right children at every node, quickly and correctly. This saves time and avoids mistakes, even for large trees.

Before vs After

✗ Before

function mirrorTree(node) {
  // manually swap left and right for each node
  if (node) {
    let temp = node.left;
    node.left = node.right;
    node.right = temp;
    mirrorTree(node.left);
    mirrorTree(node.right);
  }
}

✓ After

function mirrorTree(node) {
  if (!node) return null;
  [node.left, node.right] = [mirrorTree(node.right), mirrorTree(node.left)];
  return node;
}

What It Enables

This lets you quickly transform any binary tree into its mirror image, enabling easy visualization and solving problems that require tree symmetry.

Real Life Example

In graphics or game development, mirroring a scene graph (a tree structure) helps create reflections or symmetrical effects automatically.

Key Takeaways

Manual swapping is slow and error-prone.

Code-based mirroring swaps children recursively and correctly.

Mirroring enables quick tree transformations and symmetry checks.

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