Why Check if Two Trees are Symmetric in DSA Javascript?

The Big Idea

Discover how a simple trick can save you hours of confusing tree comparisons!

The Scenario

Imagine you have two family trees drawn on paper, and you want to see if they are mirror images of each other. Doing this by looking at every branch and comparing each node manually can be confusing and tiring.

The Problem

Manually checking each branch and node is slow and easy to mess up. You might forget to compare some parts or mix up the order, leading to wrong conclusions about whether the trees are symmetric.

The Solution

Using a simple step-by-step method, we can automatically compare the trees by checking if the left side of one tree matches the right side of the other, and vice versa. This makes the process fast, clear, and error-free.

Before vs After

✗ Before

function areSymmetric(t1, t2) {
  // Manually compare nodes one by one
  if (!t1 && !t2) return true;
  if (!t1 || !t2) return false;
  if (t1.val !== t2.val) return false;
  return areSymmetric(t1.left, t2.right) && areSymmetric(t1.right, t2.left);
}

✓ After

function isSymmetric(root) {
  function check(t1, t2) {
    if (!t1 && !t2) return true;
    if (!t1 || !t2) return false;
    if (t1.val !== t2.val) return false;
    return check(t1.left, t2.right) && check(t1.right, t2.left);
  }
  return check(root.left, root.right);
}

What It Enables

This lets us quickly and reliably tell if two trees are mirror images, which helps in many computer tasks like organizing data or checking structures.

Real Life Example

Think of checking if two family trees are mirror images to find if two families have symmetrical relationships or patterns, without drawing and comparing every branch by hand.

Key Takeaways

Manual comparison of trees is slow and error-prone.

Recursive checking of left and right nodes simplifies the process.

Automated symmetry checks save time and avoid mistakes.

Practice

1. What does it mean for two binary trees to be symmetric?

easy

A. They have the same root value only.

B. They have the same number of nodes.

C. They are mirror images of each other around their center.

D. They have identical left subtrees only.

Solution

1. Step 1: Understand symmetry in trees

   Two trees are symmetric if one is the mirror reflection of the other around the center.

2. Step 2: Compare structure and node values

   Both structure and node values must match in a mirrored way for symmetry.

3. Final Answer:

   They are mirror images of each other around their center. -> Option C

4. Quick Check:

   Symmetry means mirror image = C [OK]

Hint: Symmetric trees mirror each other exactly [OK]

Common Mistakes:

• Thinking same number of nodes means symmetric
• Only comparing root values
• Ignoring structure differences

2. Which of the following is the correct JavaScript function signature to check if two trees are symmetric?

easy

A. function checkSymmetric(tree) {}

B. function isSymmetric(tree1, tree2) {}

C. function symmetricCheck(tree1, tree2, tree3) {}

D. function isSymmetric(tree1) {}

Solution

1. Step 1: Identify parameters needed

   To check symmetry between two trees, we need two tree inputs.

2. Step 2: Match function signature

   Only the function with two parameters for two trees matches the requirement.

3. Final Answer:

   function isSymmetric(tree1, tree2) {} -> Option B

4. Quick Check:

   Two trees need two parameters = B [OK]

Hint: Two trees need two parameters in function [OK]

Common Mistakes:

• Using only one tree parameter
• Adding unnecessary third parameter
• Wrong function naming

3. What will be the output of the following code snippet?

function isMirror(t1, t2) {
  if (!t1 && !t2) return true;
  if (!t1 || !t2) return false;
  return (t1.val === t2.val) && isMirror(t1.left, t2.right) && isMirror(t1.right, t2.left);
}

const tree1 = { val: 1, left: { val: 2, left: null, right: null }, right: { val: 3, left: null, right: null } };
const tree2 = { val: 1, left: { val: 3, left: null, right: null }, right: { val: 2, left: null, right: null } };

console.log(isMirror(tree1, tree2));

medium

A. TypeError

B. false

C. undefined

D. true

Solution

1. Step 1: Check base cases for null nodes

   Both trees have matching null nodes at corresponding mirror positions, so base cases return true.

2. Step 2: Compare node values and recursive calls

   Root values match (1), left of tree1 (2) matches right of tree2 (2), right of tree1 (3) matches left of tree2 (3). All recursive calls return true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Mirror check returns true for symmetric trees [OK]

Hint: Mirror check compares opposite children recursively [OK]

Common Mistakes:

• Confusing left and right children in recursion
• Ignoring null node base cases
• Assuming order of children doesn't matter

4. Identify the bug in this code snippet that checks if two trees are symmetric:

function isMirror(t1, t2) {
  if (!t1 && !t2) return true;
  if (!t1 || !t2) return false;
  return (t1.val === t2.val) && isMirror(t1.left, t2.left) && isMirror(t1.right, t2.right);
}

medium

A. It compares left with left and right with right instead of mirror sides.

B. It does not check for null nodes.

C. It uses assignment instead of comparison.

D. It returns true too early without recursion.

Solution

1. Step 1: Analyze recursive calls for symmetry

   The function compares t1.left with t2.left and t1.right with t2.right, which checks for identical trees, not mirror symmetry.

2. Step 2: Correct recursive calls for mirror check

   To check symmetry, t1.left should be compared with t2.right and t1.right with t2.left.

3. Final Answer:

   It compares left with left and right with right instead of mirror sides. -> Option A

4. Quick Check:

   Mirror check swaps sides in recursion = A [OK]

Hint: Mirror check swaps left and right children in recursion [OK]

Common Mistakes:

• Comparing same sides instead of mirror sides
• Missing null checks
• Using assignment (=) instead of equality (===)

5. Given two binary trees, write a function to check if they are symmetric. Which approach correctly solves this problem?

hard

A. Use a queue to iteratively compare nodes from left of one tree and right of the other.

B. Compare only the root nodes of both trees.

C. Check if both trees have the same height only.

D. Traverse both trees in preorder and compare node values in order.

Solution

1. Step 1: Understand iterative symmetry check

   Using a queue, we can compare nodes from the left side of one tree with the right side of the other tree iteratively.

2. Step 2: Why other methods fail

   Preorder traversal compares nodes in the same order, not mirror order. Height check or root-only check ignores structure and values.

3. Final Answer:

   Use a queue to iteratively compare nodes from left of one tree and right of the other. -> Option A

4. Quick Check:

   Iterative mirror check uses queue swapping sides = D [OK]

Hint: Use queue to compare opposite sides iteratively [OK]

Common Mistakes:

• Using preorder traversal without mirroring
• Checking only height or root values
• Ignoring iterative approach benefits