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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Mirroring a Simple Binary Tree

What is the output of the following Go code after mirroring the binary tree? The output prints the tree nodes in preorder (root, left, right).

DSA Go

package main

import "fmt"

type Node struct {
    Val   int
    Left  *Node
    Right *Node
}

func mirror(root *Node) *Node {
    if root == nil {
        return nil
    }
    root.Left, root.Right = mirror(root.Right), mirror(root.Left)
    return root
}

func preorder(root *Node) {
    if root == nil {
        return
    }
    fmt.Print(root.Val, " ")
    preorder(root.Left)
    preorder(root.Right)
}

func main() {
    root := &Node{Val: 1}
    root.Left = &Node{Val: 2}
    root.Right = &Node{Val: 3}
    root.Left.Left = &Node{Val: 4}
    root.Left.Right = &Node{Val: 5}

    mirror(root)
    preorder(root)
}

A 1 2 3 4 5

B 1 3 2 4 5

C 1 2 3 5 4

D 1 3 2 5 4

💡 Hint

Think about swapping left and right children at every node.

Mirroring swaps left and right children recursively. The original preorder is 1 2 4 5 3. After mirroring, left and right children swap, so preorder becomes 1 3 2 5 4.

🧠 Conceptual

intermediate

Understanding Mirror Operation on Binary Trees

Which statement correctly describes the effect of the mirror operation on a binary tree?

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

B It removes all leaf nodes from the tree.

C It reverses the order of nodes in the inorder traversal only.

D It duplicates every node and attaches the duplicate as the left child.

💡 Hint

Think about what happens to each node's children after mirroring.

Mirroring a binary tree means swapping the left and right children of every node recursively. It does not remove or duplicate nodes, nor does it only affect traversal order.

🔧 Debug

advanced

Identify the Error in Mirror Function Implementation

What error does the following Go code produce when trying to mirror a binary tree?

DSA Go

func mirror(root *Node) *Node {
    if root == nil {
        return root
    }
    temp := root.Left
    root.Left = root.Right
    root.Right = temp
    mirror(root.Left)
    mirror(root.Right)
    return root
}

A Compilation error: missing return statement.

B No error; code runs correctly and mirrors the tree.

C Nil pointer dereference runtime error.

D Stack overflow due to infinite recursion.

💡 Hint

Check the base case and recursive calls carefully.

The code correctly swaps left and right children and recursively calls mirror on them. The base case returns root when nil, preventing infinite recursion. No errors occur.

❓ Predict Output

advanced

Output After Mirroring a Skewed Binary Tree

Given the following skewed binary tree and mirror function, what is the preorder output after mirroring?

DSA Go

package main

import "fmt"

type Node struct {
    Val   int
    Left  *Node
    Right *Node
}

func mirror(root *Node) *Node {
    if root == nil {
        return nil
    }
    root.Left, root.Right = mirror(root.Right), mirror(root.Left)
    return root
}

func preorder(root *Node) {
    if root == nil {
        return
    }
    fmt.Print(root.Val, " ")
    preorder(root.Left)
    preorder(root.Right)
}

func main() {
    root := &Node{Val: 1}
    root.Right = &Node{Val: 2}
    root.Right.Right = &Node{Val: 3}
    root.Right.Right.Right = &Node{Val: 4}

    mirror(root)
    preorder(root)
}

A 1 2 3 4

B 1 2 3

C 1 4 3 2

D 1 4 3

💡 Hint

Mirroring a right-skewed tree makes it left-skewed.

The original tree is skewed right: 1 -> 2 -> 3 -> 4. After mirroring, it becomes skewed left: 1 -> 4 -> 3 -> 2. Preorder prints root, left, right, so output is 1 4 3 2.

🧠 Conceptual

expert

Effect of Mirroring on Tree Height and Structure

After mirroring a binary tree, which of the following statements is always true?

A The height of the tree remains the same, but the left and right subtrees are swapped at every node.

B The height of the tree doubles because nodes are duplicated.

C The tree becomes a complete binary tree regardless of original shape.

D All leaf nodes become internal nodes after mirroring.

💡 Hint

Mirroring flips the tree horizontally but does not add or remove nodes.

Mirroring swaps left and right children at every node, preserving the height and node count. It does not duplicate nodes or change completeness or leaf status.

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 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 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 deleting nodes
• Confusing mirror with sorting
• Assuming mirror means flattening the tree

2. Which Go function signature correctly represents a function to mirror a binary tree node?

easy

A. func mirrorTree(root *TreeNode) int

B. func mirrorTree(root TreeNode) TreeNode

C. func mirrorTree(root TreeNode) bool

D. func mirrorTree(root *TreeNode) *TreeNode

Solution

1. Step 1: Understand pointer usage in Go trees

   Binary trees are usually represented with pointers to nodes to allow modification.

2. Step 2: Check return type for mirrored tree

   The function should return a pointer to the root of the mirrored tree.

3. Final Answer:

   func mirrorTree(root *TreeNode) *TreeNode -> Option D

4. Quick Check:

   Pointer input and output for tree nodes [OK]

Hint: Use pointers for tree nodes to modify and return [OK]

Common Mistakes:

• Passing tree nodes by value instead of pointer
• Returning wrong type like int or bool
• Not returning the mirrored tree root

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

type TreeNode struct {
    Val int
    Left, Right *TreeNode
}

func mirrorTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    left := mirrorTree(root.Left)
    right := mirrorTree(root.Right)
    root.Left, root.Right = right, left
    return root
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    mirrorTree(root)
    fmt.Println(root.Val, root.Left.Val, root.Right.Val)
}

medium

A. 2 1 3

B. 1 2 3

C. 1 3 2

D. 3 1 2

Solution

1. Step 1: Understand the initial tree structure

   Root is 1, left child is 2, right child is 3.

2. Step 2: Apply mirrorTree function

   Mirror swaps left and right children at root, so left becomes 3 and right becomes 2.

3. Final Answer:

   1 3 2 -> Option C

4. Quick Check:

   Swapped children values = 1 3 2 [OK]

Hint: Mirror swaps left and right children recursively [OK]

Common Mistakes:

• Not swapping children correctly
• Printing values before mirroring
• Confusing left and right child values

4. Identify the error in this Go function that attempts to mirror a binary tree:

func mirrorTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    mirrorTree(root.Left)
    mirrorTree(root.Right)
    root.Left, root.Right = root.Right, root.Left
    return root
}

medium

A. The base case is missing

B. The recursive calls do not assign the mirrored subtrees back to root.Left and root.Right

C. The function returns the wrong type

D. Swapping children should happen before recursive calls

Solution

1. Step 1: Analyze recursive calls

   The function calls mirrorTree on root.Left and root.Right but ignores their returned mirrored subtrees.

2. Step 2: Check assignment of mirrored subtrees

   Without assigning returned values, the subtrees are not updated, so mirror is incomplete.

3. Final Answer:

   The recursive calls do not assign the mirrored subtrees back to root.Left and root.Right -> Option B

4. Quick Check:

   Assign returned mirrored subtrees to children [OK]

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

Common Mistakes:

• Ignoring returned subtree pointers
• Swapping children before recursion
• Missing base case (though present here)

5. Given a binary tree, how would you modify the mirror function to also count how many nodes were swapped during the mirroring process?

hard

A. Add a counter variable and increment it each time left and right children are swapped recursively

B. Return the sum of node values instead of mirroring

C. Only swap nodes if their values are even

D. Use a queue to traverse the tree without recursion

Solution

1. Step 1: Understand the goal

   We want to count how many nodes had their children swapped during mirroring.

2. Step 2: Modify mirror function to count swaps

   Each time we swap left and right children at a node, increment a counter and add counts from recursive calls.

3. Final Answer:

   Add a counter variable and increment it each time left and right children are swapped recursively -> Option A

4. Quick Check:

   Count swaps during recursion [OK]

Hint: Count swaps inside recursion when swapping children [OK]

Common Mistakes:

• Counting node values instead of swaps
• Not accumulating counts from recursive calls
• Using traversal without swapping