Mirror a Binary Tree in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Mirroring a Simple Binary Tree

What is the output of the following TypeScript code that mirrors a binary tree and prints its inorder traversal?

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : 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 inorder(root: TreeNode | null, result: number[] = []): number[] {
  if (!root) return result;
  inorder(root.left, result);
  result.push(root.val);
  inorder(root.right, result);
  return result;
}

const root = new TreeNode(1, new TreeNode(2), new TreeNode(3));
mirrorTree(root);
console.log(inorder(root));

A [3, 1, 2]

B [2, 1, 3]

C [1, 2, 3]

D [3, 2, 1]

💡 Hint

Think about how inorder traversal visits nodes after mirroring the left and right children.

The mirrorTree function swaps the left and right children recursively. The original tree is:

1
/ \
2 3

After mirroring, it becomes:

1
/ \
3 2

Inorder traversal visits left, root, right, so the output is [3, 1, 2].

🧠 Conceptual

intermediate

Understanding Mirror Tree Recursive Calls

In the mirrorTree function, what is the main purpose of the recursive calls before swapping the children?

A To swap the left and right children of the current node only

B To print the values of the nodes in preorder

C To count the number of nodes in the tree

D To ensure the entire left and right subtrees are mirrored before swapping

💡 Hint

Think about what happens if you swap children before mirroring subtrees.

The recursive calls mirror the left and right subtrees fully before swapping them at the current node. This ensures the entire tree is mirrored correctly.

🔧 Debug

advanced

Identify the Bug in Mirror Tree Implementation

What error will this code produce when trying to mirror a binary tree?

DSA Typescript

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;
}

const root = new TreeNode(1, new TreeNode(2), new TreeNode(3));
mirrorTree(root);
console.log(root.left?.val);

A No error; output is 3

B TypeError because root.left or root.right is undefined

C Stack overflow due to infinite recursion

D Output is 2 instead of 3

💡 Hint

Check if the recursive calls are used correctly and if swapping is done properly.

The code correctly mirrors the tree by recursively calling mirrorTree on left and right subtrees, then swapping them. The output root.left?.val is 3 after mirroring.

❓ Predict Output

advanced

Output of Mirroring a Larger Binary Tree

What is the output of the inorder traversal after mirroring this binary tree?

DSA Typescript

const root = new TreeNode(4,
  new TreeNode(2, new TreeNode(1), new TreeNode(3)),
  new TreeNode(7, new TreeNode(6), new TreeNode(9))
);
mirrorTree(root);
console.log(inorder(root));

A [1, 2, 3, 4, 6, 7, 9]

B [9, 7, 6, 4, 3, 2, 1]

C [6, 7, 9, 4, 1, 2, 3]

D [3, 2, 1, 4, 9, 7, 6]

💡 Hint

Remember mirroring swaps left and right subtrees at every node.

The original inorder traversal is [1, 2, 3, 4, 6, 7, 9]. After mirroring, the tree's left and right children are swapped at every node, reversing the inorder traversal to [9, 7, 6, 4, 3, 2, 1].

🧠 Conceptual

expert

Time Complexity of Mirroring a Binary Tree

What is the time complexity of the mirrorTree function for a binary tree with n nodes?

A O(log n), because the tree is divided in half each recursion

B O(n^2), because each node swaps children multiple times

C O(n), because each node is visited once

D O(1), because swapping is done in constant time

💡 Hint

Consider how many times the function visits each node.

The mirrorTree function visits each node exactly once to swap its children, so the time complexity is O(n).

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