Mirror a Binary Tree in DSA Javascript - Execution Trace

Concept Flow - Mirror a Binary Tree

Start at root node

↓

Swap left and right children

↓

Mirror left subtree

↓

Mirror right subtree

↓

Return mirrored node

↓

Done

The process starts at the root, swaps its children, then recursively mirrors left and right subtrees until all nodes are processed.

Execution Sample

DSA Javascript

function mirror(node) {
  if (!node) return null;
  [node.left, node.right] = [node.right, node.left];
  mirror(node.left);
  mirror(node.right);
  return node;
}

This function swaps left and right children of each node recursively to create the mirror image of the binary tree.

Execution Table

Step	Operation	Node Visited	Pointer Changes	Tree State
1	Start at root	1	None	1 / \ 2 3
2	Swap children	1	left=3, right=2	1 / \ 3 2
3	Mirror left subtree	3	None	1 / \ 3 2
4	Swap children	3	left=6, right=5	1 / \ 3 2 / \ 6 5
5	Mirror left subtree	6	None	1 / \ 3 2 / \ 6 5
6	Swap children	6	left=null, right=null	1 / \ 3 2 / \ 6 5
7	Mirror left subtree	null	None	No change
8	Mirror right subtree	null	None	No change
9	Mirror right subtree	5	None	1 / \ 3 2 / \ 6 5
10	Swap children	5	left=null, right=null	1 / \ 3 2 / \ 6 5
11	Mirror left subtree	null	None	No change
12	Mirror right subtree	null	None	No change
13	Mirror right subtree	2	None	1 / \ 3 2
14	Swap children	2	left=4, right=null	1 / \ 3 2 / 4 null
15	Mirror left subtree	4	None	1 / \ 3 2 / 4 null
16	Swap children	4	left=null, right=null	1 / \ 3 2 / 4 null
17	Mirror left subtree	null	None	No change
18	Mirror right subtree	null	None	No change
19	Mirror right subtree	null	None	No change
20	Return mirrored root	1	None	1 / \ 3 2 / \ / 6 5 4 null
21	Done	None	None	Final mirrored tree

💡 All nodes visited and children swapped, recursion ends.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 10	After Step 14	Final
node	root(1)	root(1)	node(3)	node(6)	node(5)	node(2)	root(1)
node.left	2	3	6	null	null	4	3
node.right	3	2	5	null	null	null	2
Tree State	1 / \ 2 3	1 / \ 3 2	1 / \ 3 2 / \ 6 5	No change	No change	1 / \ 3 2 / 4 null	1 / \ 3 2 / \ / 6 5 4 null

Key Moments - 3 Insights

Why do we swap children before recursive calls?

Swapping children first (see Step 2) ensures that when recursion goes to node.left and node.right, it processes the mirrored positions, so the entire subtree gets mirrored correctly.

What happens when a node is null?

When node is null (Steps 7, 8, 11, 12, 17, 18, 19), the function returns immediately without changes, stopping recursion on that path.

Why does the tree state update after swapping children?

Because swapping changes the structure visually (Step 2, 4, 10, 14), the tree state reflects the new left and right children positions after each swap.

Visual Quiz - 3 Questions

Look at the execution table, what are the left and right children of node 1 after Step 2?

A left=5, right=6

B left=2, right=3

C left=3, right=2

D left=null, right=null

💡 Hint

Check the 'Pointer Changes' column at Step 2 in the execution table.

At which step does the node 2 swap its children?

A Step 10

B Step 14

C Step 4

D Step 6

💡 Hint

Look for node 2 in the 'Node Visited' column and find when 'Swap children' happens.

If we skip swapping children before recursion, how would the tree state after Step 2 change?

A Children of node 1 remain as left=2, right=3

B Children of node 1 become left=3, right=2

C Tree becomes empty

D Children of node 1 become null

💡 Hint

Refer to the 'Pointer Changes' column at Step 2 and understand the effect of swapping.

Concept Snapshot

Mirror a Binary Tree:
- Swap left and right children of each node
- Recursively mirror left subtree
- Recursively mirror right subtree
- Base case: if node is null, return
- Result: tree is flipped horizontally

Full Transcript

To mirror a binary tree, start at the root node. Swap its left and right children. Then recursively apply the same process to the left and right subtrees. If a node is null, return immediately. This process flips the tree horizontally, creating a mirror image. The execution table shows each step visiting nodes, swapping children, and updating the tree structure until all nodes are processed.

Practice

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

easy

A. Sort the nodes in ascending order

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

C. Convert the tree into a linked list

D. Remove all leaf nodes from the tree

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 swapping 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 is the correct JavaScript syntax to swap the left and right children of a node named node?

easy

A. let temp = node.left; node.left = node.right; node.right = temp;

B. node.left = node.right; node.right = node.left;

C. node.left, node.right = node.right, node.left;

D. swap(node.left, node.right);

Solution

1. Step 1: Understand swapping variables in JavaScript

   To swap two variables, use a temporary variable to hold one value during the swap.

2. Step 2: Check each option

   let temp = node.left; node.left = node.right; node.right = temp; correctly uses a temporary variable to swap node.left and node.right. node.left = node.right; node.right = node.left; overwrites node.left before assigning node.right. node.left, node.right = node.right, node.left; uses invalid syntax in JavaScript. swap(node.left, node.right); calls an undefined function.

3. Final Answer:

   let temp = node.left; node.left = node.right; node.right = temp; -> Option A

4. Quick Check:

   Use temp variable to swap [OK]

Hint: Use a temp variable to swap two values in JavaScript [OK]

Common Mistakes:

• Overwriting one child before saving it
• Using Python-style tuple swap syntax
• Calling undefined swap functions

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

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function mirror(node) {
  if (!node) return null;
  let left = mirror(node.left);
  let right = mirror(node.right);
  node.left = right;
  node.right = left;
  return node;
}

// Tree before mirroring:
//     1
//    / \
//   2   3
//  /     \
// 4       5

let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.right.right = new Node(5);

mirror(root);

function printTree(node) {
  if (!node) return 'null';
  return node.val + ' -> ' + printTree(node.left) + ', ' + printTree(node.right);
}

console.log(printTree(root));

medium

A. 1 -> 2 -> 4 -> null, 3 -> 5 -> null, null

B. 1 -> 3 -> 4 -> null, 2 -> 5 -> null, null

C. 1 -> 3 -> null, 2 -> 5 -> null, 4

D. 1 -> 3 -> 5 -> null, 2 -> 4 -> null, null

Solution

1. Step 1: Understand the mirror function

   The mirror function swaps left and right children recursively for every node.

2. Step 2: Trace the tree after mirroring

   Original tree: 1 with left=2 and right=3; 2 has left=4; 3 has right=5.
   After mirroring:
   1's left becomes 3, right becomes 2.
   3's left becomes 5, right becomes null.
   2's left becomes null, right becomes 4.
   Printing in preorder with left then right shows: 1 -> 3 -> 5 -> null, 2 -> 4 -> null, null.

3. Final Answer:

   1 -> 3 -> 5 -> null, 2 -> 4 -> null, null -> Option D

4. Quick Check:

   Mirror swaps children recursively [OK]

Hint: Trace swaps at each node recursively [OK]

Common Mistakes:

• Not swapping children at all nodes
• Mixing left and right children incorrectly
• Misreading the print output format

4. Find the error in the following mirror function and select the correct fix:

function mirror(node) {
  if (node == null) return;
  mirror(node.left);
  mirror(node.right);
  let temp = node.left;
  node.left = node.right;
  node.right = temp;
}

medium

A. The function should return the node after swapping children

B. The base case should check for undefined instead of null

C. The swap should happen before recursive calls

D. The temp variable should be declared outside the function

Solution

1. Step 1: Analyze the function behavior

   The function swaps children after recursive calls but does not return the node, so the caller gets undefined.

2. Step 2: Identify the fix

   Returning the node after swapping ensures the mirrored subtree is returned properly for further use.

3. Final Answer:

   The function should return the node after swapping children -> Option A

4. Quick Check:

   Return node after swap [OK]

Hint: Always return the node after modifying it in recursion [OK]

Common Mistakes:

• Not returning the node causes undefined results
• Swapping before recursion breaks logic
• Misunderstanding null vs undefined checks

5. Given a binary tree where some nodes have only one child, how does the mirror function handle these nodes?
Consider the tree:

    10
   /  
  20   
    \
     30

What will be the structure after mirroring?

hard

A. 10 with left child 30 and right child 20; 20 and 30 have no children

B. 10 with left child 20 and right null; 20 with left null and right 30

C. 10 with left child null and right child 20; 20 with left child 30 and right null

D. 10 with left child null and right child 30; 30 with left child 20 and right null

Solution

1. Step 1: Understand mirroring nodes with single children

   Mirroring swaps left and right children even if one child is null.

2. Step 2: Apply mirror to given tree

   Original:
   10 has left=20, right=null
   20 has left=null, right=30
   After mirror:
   10's left becomes null, right becomes 20
   20's left becomes 30, right becomes null

3. Final Answer:

   10 with left child null and right child 20; 20 with left child 30 and right null -> Option C

4. Quick Check:

   Mirror swaps children even if one is null [OK]

Hint: Swap children even if one child is missing [OK]

Common Mistakes:

• Ignoring null children during swap
• Assuming children move to wrong side
• Confusing node values with positions