Why Mirror a Binary Tree in DSA Typescript?

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 leaf on the opposite side, which is tiring and easy to mess up.

The Problem

Manually flipping each branch and leaf is slow and prone to mistakes. You might forget to swap some parts or place them incorrectly, making the mirrored tree wrong and confusing.

The Solution

Using a simple method to swap the left and right branches of every node in the tree automatically creates the mirror image. This saves time and ensures accuracy by handling every part systematically.

Before vs After

✗ Before

function mirrorTree(node) {
  // manually redraw each branch
  // very complex and error-prone
}

✓ 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 create a mirrored version of any tree, useful in graphics, data transformations, and problem-solving.

Real Life Example

In photo editing apps, flipping images horizontally is like mirroring a tree structure of pixels, making the picture look reversed instantly.

Key Takeaways

Manually mirroring a tree is slow and error-prone.

Swapping left and right children recursively automates the mirror process.

This method is fast, reliable, and easy to implement.

Practice

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

easy

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

B. Remove all leaf nodes from the tree

C. Convert the tree into a linked list

D. Sort the tree 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: Effect on the tree structure

   This operation flips the tree around its center, creating a mirror image.

3. Final Answer:

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

4. Quick Check:

   Mirror = Swap children [OK]

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

Common Mistakes:

• Thinking mirror means sorting nodes
• Confusing mirror with removing nodes
• Assuming mirror changes node values

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

easy

A. function mirrorTree(root: string): string

B. function mirrorTree(root: number): number

C. function mirrorTree(root: TreeNode): void

D. function mirrorTree(root: TreeNode | null): TreeNode | null

Solution

1. Step 1: Identify the input and output types

   The function should accept a TreeNode or null and return a TreeNode or null after mirroring.

2. Step 2: Check each option's signature

   function mirrorTree(root: TreeNode | null): TreeNode | null matches the expected input and output types for a binary tree mirror function.

3. Final Answer:

   function mirrorTree(root: TreeNode | null): TreeNode | null -> Option D

4. Quick Check:

   Correct signature uses TreeNode or null [OK]

Hint: Mirror function inputs and outputs TreeNode or null [OK]

Common Mistakes:

• Using primitive types like number or string instead of TreeNode
• Returning void instead of the mirrored tree
• Not allowing null as input

3. Given the following TypeScript code, what is the output when the tree is mirrored and then printed in preorder?

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function mirrorTree(root: TreeNode | null): TreeNode | null {
  if (!root) return null;
  const left = mirrorTree(root.left);
  const right = mirrorTree(root.right);
  root.left = right;
  root.right = left;
  return root;
}

function preorder(root: TreeNode | null): number[] {
  if (!root) return [];
  return [root.val, ...preorder(root.left), ...preorder(root.right)];
}

const tree = new TreeNode(1, new TreeNode(2, new TreeNode(4), new TreeNode(5)), new TreeNode(3));
const mirrored = mirrorTree(tree);
console.log(preorder(mirrored));

medium

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

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

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

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

Solution

1. Step 1: Understand the original tree structure

   The tree is:
   1
   ├─ 2
   │ ├─ 4
   │ └─ 5
   └─ 3

2. Step 2: Mirror the tree by swapping left and right children

   After mirroring:
   1
   ├─ 3
   └─ 2
   ├─ 5
   └─ 4

3. Step 3: Preorder traversal of mirrored tree

   Preorder visits root, left, right:
   1, 3, 2, 5, 4

4. Final Answer:

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

5. Quick Check:

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

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

Common Mistakes:

• Not swapping children correctly
• Confusing preorder with inorder or postorder
• Forgetting to mirror recursively

4. The following TypeScript function is intended to mirror a binary tree but contains a bug. What is the bug?

function mirrorTree(root: TreeNode | null): TreeNode | null {
  if (!root) return null;
  mirrorTree(root.left);
  mirrorTree(root.right);
  const temp = root.left;
  root.left = root.right;
  root.right = temp;
  return root;
}

medium

A. The base case should return root instead of null

B. The swapping of left and right children is done before recursion

C. The recursive calls do not assign the mirrored subtrees back to root.left and root.right

D. The function does not handle empty trees

Solution

1. Step 1: Analyze the recursive calls

   The function calls mirrorTree on root.left and root.right but does not assign the results back to root.left and root.right.

2. Step 2: Understand the effect of missing assignments

   Without assigning, the subtrees are not updated with their mirrored versions, so the mirror is incomplete.

3. Final Answer:

   The recursive calls do not assign the mirrored subtrees back to root.left and root.right -> Option C

4. Quick Check:

   Assign recursive results to children [OK]

Hint: Always assign recursive mirror results back to children [OK]

Common Mistakes:

• Ignoring return values of recursive calls
• Swapping children before recursion
• Incorrect base case handling

5. You have a binary tree where some nodes have only one child (either left or right). 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. Nodes with only left child will have only right child after mirroring

B. Nodes with only left child will have no children after mirroring

C. Nodes with only right child will have both children after mirroring

D. The tree structure remains unchanged after mirroring

Solution

1. Step 1: Understand mirroring effect on single-child nodes

   Mirroring swaps left and right children of every node.

2. Step 2: Effect on nodes with only one child

   If a node has only a left child, after mirroring, that child becomes the right child, and left becomes null.

3. Final Answer:

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

4. Quick Check:

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

Hint: Mirror swaps children, single left child moves to right [OK]

Common Mistakes:

• Assuming children disappear after mirroring
• Thinking mirroring adds children
• Believing tree structure stays same