Mirror a Binary Tree in DSA Go - Build from Scratch

Mirror a Binary Tree

📖 Scenario: You are working on a simple program that deals with binary trees. A binary tree is like a family tree where each person has up to two children: a left child and a right child.Sometimes, you want to flip this tree so that the left and right children of every node swap places. This is called mirroring the tree.

🎯 Goal: Build a Go program that creates a binary tree, then mirrors it by swapping left and right children at every node, and finally prints the mirrored tree in a simple format.

📋 What You'll Learn

Create a binary tree node struct with integer values and left/right pointers

Build a specific binary tree with given values

Write a function to mirror the binary tree by swapping left and right children recursively

Print the tree nodes in preorder traversal after mirroring

💡 Why This Matters

🌍 Real World

Mirroring trees is useful in graphics, image processing, and data structure transformations.

💼 Career

Understanding tree manipulations and recursion is essential for software engineering roles involving data structures and algorithms.

1

Create the binary tree structure and build the tree

Define a struct called Node with fields Val int, Left *Node, and Right *Node. Then create a variable root that builds this exact tree:

1
/ \\
2 3
/ / \\
4 5 6

Use &Node{Val: value, Left: ..., Right: ...} syntax to build the tree.

DSA Go

// Define Node struct and create root node with given structure
// Your code here

Hint

Start by defining the Node struct with the three fields. Then build the tree from the bottom up using nested &Node{} expressions.

2

Add a helper function to mirror the tree

Create a function called mirror that takes a pointer to Node named root. This function should swap the Left and Right children of root and then call itself recursively on both children if they are not nil.

DSA Go

package main

import "fmt"

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

func mirror(root *Node) {
    // Your code here
}

func main() {
    root := &Node{
        Val: 1,
        Left: &Node{
            Val: 2,
            Left: &Node{Val: 4},
            Right: nil,
        },
        Right: &Node{
            Val: 3,
            Left:  &Node{Val: 5},
            Right: &Node{Val: 6},
        },
    }
}

Hint

Check if root is nil. If not, swap Left and Right. Then call mirror recursively on both children.

3

Call the mirror function on the tree

In the main function, call the mirror function with the root node to mirror the entire tree.

DSA Go

package main

import "fmt"

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

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

func main() {
    root := &Node{
        Val: 1,
        Left: &Node{
            Val: 2,
            Left: &Node{Val: 4},
            Right: nil,
        },
        Right: &Node{
            Val: 3,
            Left:  &Node{Val: 5},
            Right: &Node{Val: 6},
        },
    }
    // Call mirror function here
    // Your code here
}

Hint

Simply call mirror(root) inside main after building the tree.

4

Print the mirrored tree in preorder traversal

Create a function called printPreorder that takes a pointer to Node named root. It should print the Val of each node in preorder (root, left, right) separated by spaces. Then call printPreorder(root) in main after mirroring the tree.

DSA Go

package main

import "fmt"

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

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

func printPreorder(root *Node) {
    // Your code here
}

func main() {
    root := &Node{
        Val: 1,
        Left: &Node{
            Val: 2,
            Left: &Node{Val: 4},
            Right: nil,
        },
        Right: &Node{
            Val: 3,
            Left:  &Node{Val: 5},
            Right: &Node{Val: 6},
        },
    }
    mirror(root)
    // Call printPreorder here
    // Your code here
}

Hint

Print the current node's value, then recursively print the left subtree, then the right subtree. Call this function after mirroring.

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