Check if Two Trees are Symmetric in DSA Javascript - Time & Space Complexity

Time Complexity: Check if Two Trees are Symmetric

O(n)

Understanding Time Complexity

We want to know how the time needed to check if two trees are mirror images grows as the trees get bigger.

How does the number of steps change when the trees have more nodes?

Scenario Under Consideration

Analyze the time complexity 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);
}

function isSymmetric(root) {
  return isMirror(root, root);
}
    

This code checks if a tree is symmetric by comparing it to itself in a mirrored way.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls comparing pairs of nodes.
• How many times: Each node in the tree is visited once in the recursion.

How Execution Grows With Input

As the tree grows, the function checks more pairs of nodes, roughly one pair per node.

Input Size (n)	Approx. Operations
10	About 10 node comparisons
100	About 100 node comparisons
1000	About 1000 node comparisons

Pattern observation: The number of steps grows roughly in direct proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check symmetry grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The function only checks half the nodes, so it runs in half the time."

[OK] Correct: Even though it compares pairs, every node is still visited once, so the time depends on all nodes.

Interview Connect

Understanding how recursion explores all nodes helps you explain your solution clearly and confidently in interviews.

Self-Check

"What if we changed the code to check symmetry only on one subtree instead of both? How would the time complexity change?"

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