Mirror a Binary Tree in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap the left and right child nodes of the root.

DSA Go

func mirror(root *Node) {
    if root == nil {
        return
    }
    root.Left, root.Right = root.Right, root.[1]
}

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

ALeft

BRight

CParent

DChild

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping root.Right with root.Right instead of root.Left.

Using incorrect field names like Parent or Child.

Swapping root.Left and root.Right requires accessing root.Left on the right side to complete the swap.

2fill in blank

medium

Complete the code to recursively mirror the left subtree.

DSA Go

func mirror(root *Node) {
    if root == nil {
        return
    }
    mirror(root.[1])
    mirror(root.Right)
    root.Left, root.Right = root.Right, root.Left
}

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

AChild

BRight

CParent

DLeft

Attempts:

3 left

💡 Hint

Common Mistakes

Calling mirror on root.Right twice.

Using incorrect field names.

To mirror the left subtree, call mirror on root.Left recursively.

3fill in blank

hard

Fix the error in the recursive call to mirror the right subtree.

DSA Go

func mirror(root *Node) {
    if root == nil {
        return
    }
    mirror(root.Left)
    mirror(root.[1])
    root.Left, root.Right = root.Right, root.Left
}

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

ARight

BLeft

CParent

DChild

Attempts:

3 left

💡 Hint

Common Mistakes

Calling mirror on root.Left twice.

Using incorrect field names.

To mirror the right subtree, call mirror on root.Right recursively.

4fill in blank

hard

Fill both blanks to complete the mirror function that recursively mirrors the tree.

DSA Go

func mirror(root *Node) {
    if root == nil {
        return
    }
    mirror(root.[1])
    mirror(root.[2])
    root.Left, root.Right = root.Right, root.Left
}

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

ALeft

BRight

CParent

DChild

Attempts:

3 left

💡 Hint

Common Mistakes

Calling mirror on the same child twice.

Swapping children before recursive calls.

The function must recursively call mirror on both left and right children before swapping them.

5fill in blank

hard

Fill all three blanks to implement a complete mirror function with base case, recursive calls, and swap.

DSA Go

func mirror(root *Node) {
    if root == [1] {
        return
    }
    mirror(root.[2])
    mirror(root.[3])
    root.Left, root.Right = root.Right, root.Left
}

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

Anil

BLeft

CRight

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect base case condition.

Swapping children before recursive calls.

The base case checks if root is nil, then recursively mirror left and right children before swapping.

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