Check if Two Trees are Symmetric in DSA Typescript - 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 work increase when the trees have more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function isSymmetric(t1: TreeNode | null, t2: TreeNode | null): boolean {
  if (!t1 && !t2) return true;
  if (!t1 || !t2) return false;
  return (t1.val === t2.val) &&
         isSymmetric(t1.left, t2.right) &&
         isSymmetric(t1.right, t2.left);
}
    

This function checks if two trees are mirror images by comparing nodes recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls comparing pairs of nodes.
• How many times: Once for each node in the trees, visiting all nodes.

How Execution Grows With Input

As the number of nodes grows, the function checks each node pair once.

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

Pattern observation: The work grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check symmetry grows in a straight line with the number of nodes.

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, it still visits every node once, so the time grows with all nodes.

Interview Connect

Understanding how recursive tree checks scale helps you explain your approach clearly and confidently in interviews.

Self-Check

"What if we changed the function to check symmetry iteratively using a queue? How would the time complexity change?"

Practice

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

easy

A. They are mirror images of each other around their root.

B. They have the same number of nodes.

C. They have the same values in the same order.

D. They are both balanced binary search trees.

Solution

1. Step 1: Understand symmetry in trees

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

2. Step 2: Differentiate from other properties

   Having the same number of nodes or values in order does not guarantee symmetry; symmetry requires mirrored structure and values.

3. Final Answer:

   They are mirror images of each other around their root. -> Option A

4. Quick Check:

   Symmetry means mirror image = A [OK]

Hint: Symmetric trees mirror each other around the root [OK]

Common Mistakes:

• Confusing symmetry with identical trees
• Thinking same node count means symmetry
• Assuming balanced trees are always symmetric

2. Which TypeScript function signature correctly represents a function to check if two trees are symmetric?

easy

A. function symmetricCheck(root1: TreeNode, root2: TreeNode): void

B. function isSymmetric(root1: TreeNode, root2: TreeNode): boolean

C. function checkSymmetry(tree1: TreeNode, tree2: TreeNode): number

D. function isSymmetric(root: TreeNode): boolean

Solution

1. Step 1: Identify correct parameters and return type

   The function should take two tree roots and return a boolean indicating symmetry.

2. Step 2: Check function signature correctness

   function isSymmetric(root1: TreeNode, root2: TreeNode): boolean matches this: two TreeNode parameters and boolean return type.

3. Final Answer:

   function isSymmetric(root1: TreeNode, root2: TreeNode): boolean -> Option B

4. Quick Check:

   Two roots + boolean return = D [OK]

Hint: Symmetry check needs two roots and boolean output [OK]

Common Mistakes:

• Using only one root parameter
• Returning number instead of boolean
• Using void return type

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

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function isMirror(t1: TreeNode | null, t2: TreeNode | null): boolean {
  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 = new TreeNode(1, new TreeNode(2), new TreeNode(3));
const tree2 = new TreeNode(1, new TreeNode(3), new TreeNode(2));
console.log(isMirror(tree1, tree2));

medium

A. SyntaxError

B. false

C. true

D. RuntimeError

Solution

1. Step 1: Understand the isMirror function logic

   It checks if two trees are mirror images by comparing root values and recursively checking left of one with right of other and vice versa.

2. Step 2: Trace the given trees

   tree1 has root 1, left 2, right 3; tree2 has root 1, left 3, right 2. They are mirrors, so isMirror returns true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   Mirror check returns true for these trees [OK]

Hint: Mirror check compares opposite children recursively [OK]

Common Mistakes:

• Confusing left and right child comparison
• Assuming same structure means mirror
• Ignoring null checks

4. Identify the bug in this TypeScript function meant to check if two trees are symmetric:

function isSymmetric(t1: TreeNode | null, t2: TreeNode | null): boolean {
  if (!t1 && !t2) return true;
  if (!t1 || !t2) return false;
  if (t1.val !== t2.val) return false;
  return isSymmetric(t1.left, t2.left) && isSymmetric(t1.right, t2.right);
}

medium

A. It does not check node values before recursion.

B. It misses the base case when both nodes are null.

C. It returns true when one node is null and the other is not.

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

Solution

1. Step 1: Analyze recursive calls

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

2. Step 2: Identify correct mirror comparison

   For 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 D

4. Quick Check:

   Mirror needs cross child comparison = A [OK]

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

Common Mistakes:

• Comparing same side children instead of opposite
• Forgetting to check node values before recursion
• Missing null base cases

5. Given two binary trees, write a TypeScript function that returns true if they are symmetric around their roots, and false otherwise. Which approach correctly solves this problem?

hard

A. Use a helper function to recursively compare left subtree of one tree with right subtree of the other and vice versa.

B. Traverse both trees in preorder and compare the traversal arrays for equality.

C. Check if both trees have the same height and node count.

D. Compare only the root node values of both trees.

Solution

1. Step 1: Understand symmetry check requirements

   Symmetry means the left subtree of one tree is the mirror of the right subtree of the other.

2. Step 2: Choose correct approach

   Recursively comparing left subtree of one with right subtree of the other ensures mirror symmetry is checked properly.

3. Final Answer:

   Use a helper function to recursively compare left subtree of one tree with right subtree of the other and vice versa. -> Option A

4. Quick Check:

   Mirror recursion approach = B [OK]

Hint: Mirror symmetry needs cross subtree recursive comparison [OK]

Common Mistakes:

• Comparing preorder traversals which miss mirror structure
• Checking only height or node count
• Comparing only root values