Mirror a Binary Tree in DSA Java - Build from Scratch

Mirror a Binary Tree

📖 Scenario: You are working with a simple binary tree structure used in many applications like file systems and expression parsing. Sometimes, you need to create a mirror image of this tree, where the left and right children of all nodes are swapped.

🎯 Goal: Build a Java 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 order.

📋 What You'll Learn

Create a TreeNode class with int val, TreeNode left, and TreeNode right fields

Create a binary tree with the exact structure given

Write a method mirror(TreeNode root) that swaps left and right children recursively

Write an inOrder(TreeNode root) method to print the tree nodes in order

Print the in-order traversal of the mirrored tree

💡 Why This Matters

🌍 Real World

Mirroring trees is useful in graphics, image processing, and reversing hierarchical data structures.

💼 Career

Understanding tree manipulation is important for software engineers working with data structures, algorithms, and system design.

1

Create the TreeNode class and build the initial binary tree

Create a class called TreeNode with an integer field val, and two TreeNode fields called left and right. Then create a binary tree with root node value 1, left child value 2, and right child value 3. The left child of node 2 should be 4 and the right child should be 5. Assign this tree to a variable called root.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

// Your code here

Hint

Define the TreeNode class first. Then create the root and assign its children exactly as described.

2

Add the mirror method to swap left and right children

Add a static method called mirror that takes a TreeNode root parameter. This method should swap the left and right children of the root node, then recursively call itself on the left and right children if they are not null.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

public class Main {
    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);
    }

    // Add the mirror method below
    // Your code here
}

Hint

Use a temporary variable to swap root.left and root.right. Then call mirror recursively on both children.

3

Add an inOrder method to print the tree nodes in order

Add a static method called inOrder that takes a TreeNode root parameter. This method should print the values of the tree nodes in in-order traversal (left, root, right) separated by spaces. Use recursion and print each value followed by a space without a newline.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

public class Main {
    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);
    }

    public static void mirror(TreeNode root) {
        if (root == null) {
            return;
        }
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;

        mirror(root.left);
        mirror(root.right);
    }

    // Add the inOrder method below
    // Your code here
}

Hint

Use recursion to visit left child, print current node value, then visit right child.

4

Call mirror and print the mirrored tree in order

In the main method, after building the tree, call mirror(root) to mirror the tree. Then call inOrder(root) to print the mirrored tree nodes in order on a single line.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

public class Main {
    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);

        // Call mirror and inOrder here
        // Your code here
    }

    public static void mirror(TreeNode root) {
        if (root == null) {
            return;
        }
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;

        mirror(root.left);
        mirror(root.right);
    }

    public static void inOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        inOrder(root.left);
        System.out.print(root.val + " ");
        inOrder(root.right);
    }
}

Hint

Call mirror(root) first, then inOrder(root) to print the mirrored tree.

Practice

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

easy

A. Remove all leaf nodes from the tree

B. Swap the left and right children of every node in 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 B

4. Quick Check:

   Mirror = Swap children [OK]

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

Common Mistakes:

• Thinking mirror means deleting nodes
• Confusing mirror with sorting
• Assuming mirror flattens the tree

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

easy

A. public void mirrorTree(Node root)

B. public int mirrorTree(Node root)

C. void mirrorTree(Node root) throws Exception

D. public Node mirrorTree()

Solution

1. Step 1: Identify correct return type

   Mirroring modifies the tree in place, so void return type is appropriate.

2. Step 2: Check method parameters

   The method should accept the root node to start mirroring from.

3. Final Answer:

   public void mirrorTree(Node root) -> Option A

4. Quick Check:

   Void return + root parameter = correct signature [OK]

Hint: Mirror modifies tree in place, so use void with root parameter [OK]

Common Mistakes:

• Using return type int instead of void
• Missing root parameter
• Adding unnecessary exceptions

3. Given the following binary tree and the mirror function below, what will be the preorder traversal output after mirroring?

    1
   / \
  2   3
 /     \
4       5

void mirrorTree(Node root) {
  if (root == null) return;
  Node temp = root.left;
  root.left = root.right;
  root.right = temp;
  mirrorTree(root.left);
  mirrorTree(root.right);
}

medium

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

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

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

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

Solution

1. Step 1: Understand the mirror function

   The function swaps left and right children at every node recursively.

2. Step 2: Apply mirror to the tree

   Original tree preorder: 1,2,4,3,5. After mirror, left and right children swap: - Root 1: left=3, right=2 - Node 3: left=5, right=null - Node 2: left=null, right=4

3. Step 3: Preorder traversal after mirror

   Visit root, then left subtree, then right subtree: 1, 3, 5, 2, 4

4. Final Answer:

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

5. Quick Check:

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

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

Common Mistakes:

• Not swapping children correctly
• Confusing preorder with inorder
• Missing recursive calls

4. The following code attempts to mirror a binary tree but causes a NullPointerException. What is the error?

public void mirrorTree(Node root) {
  Node temp = root.left;
  root.left = root.right;
  root.right = temp;
  mirrorTree(root.left);
  mirrorTree(root.right);
}

medium

A. Swapping left and right children incorrectly

B. Recursive calls should be after swapping

C. Missing null check before accessing root.left

D. Method should return a Node instead of void

Solution

1. Step 1: Identify cause of NullPointerException

   The code accesses root.left without checking if root is null.

2. Step 2: Fix by adding null check

   Adding "if (root == null) return;" at the start prevents errors when root is null.

3. Final Answer:

   Missing null check before accessing root.left -> Option C

4. Quick Check:

   NullPointerException = missing null check [OK]

Hint: Always check if node is null before accessing children [OK]

Common Mistakes:

• Ignoring null checks
• Thinking swap order causes error
• Changing return type unnecessarily

5. You have a binary tree where some nodes have only one child. 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. Leaf nodes become internal nodes after mirroring

B. Nodes with one child will have two children after mirroring

C. The tree height will always increase by 1

D. Nodes with only left child will have only right child after mirroring

Solution

1. Step 1: Understand mirroring effect on children

   Mirroring swaps left and right children at every node.

2. Step 2: Analyze nodes with one child

   If a node has only a left child, after mirroring it will have only a right child (the left child moves to right).

3. Step 3: Check other options

   Tree height does not necessarily increase; leaf nodes remain leaves; nodes do not gain extra children.

4. Final Answer:

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

5. Quick Check:

   Mirror swaps children, so left-only becomes right-only [OK]

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

Common Mistakes:

• Assuming tree height changes
• Thinking nodes gain extra children
• Confusing leaf and internal nodes