Check if Two Trees are Symmetric in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

Two binary trees are symmetric if one is the mirror image of the other. This means the left subtree of one tree is the mirror of the right subtree of the other tree, and vice versa.

beginner

What is the base case when checking if two trees are symmetric?

If both trees are empty (NULL), they are symmetric. If only one is empty, they are not symmetric.

intermediate

Which recursive checks are needed to confirm two trees are symmetric?

Check if the root values are equal, then recursively check if the left subtree of the first tree is symmetric with the right subtree of the second tree, and the right subtree of the first tree is symmetric with the left subtree of the second tree.

beginner

Why do we compare left subtree of one tree with right subtree of the other tree?

Because symmetry means mirror image, so the left side of one tree should match the right side of the other tree to be symmetric.

intermediate

What is the time complexity of checking if two trees are symmetric?

The time complexity is O(n), where n is the number of nodes in the trees, because each node is visited once during the comparison.

What should be returned if both input trees are NULL when checking symmetry?

A True

B False

C Depends on tree size

D Error

Two empty trees are symmetric by definition.

When checking symmetry, which subtree pairs are compared?

A Left with left, right with right

B No subtree comparison needed

C Left with right, right with left

D Root with root only

Symmetry means mirror image, so left subtree of one tree is compared with right subtree of the other.

If the root values of two trees differ, what can we conclude?

A Trees are identical

B Trees are symmetric

C Check subtrees to decide

D Trees are not symmetric

Root values must be equal for trees to be symmetric.

What is the main technique used to check if two trees are symmetric?

A Iterative traversal

B Recursion

C Sorting nodes

D Hashing

Recursion is used to compare subtrees in a mirror fashion.

What is the worst-case time complexity of checking symmetry for two trees with n nodes?

A O(n)

B O(log n)

C O(1)

D O(n^2)

Each node is visited once, so time complexity is linear O(n).

Explain how to check if two binary trees are symmetric using recursion.

Think about mirror images and how left and right subtrees relate.

Describe why comparing left subtree of one tree with right subtree of the other is necessary for symmetry check.

Imagine folding one tree over the other.

Practice

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

easy

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

B. They have the same number of nodes

C. They have the same preorder traversal

D. They have the same height

Solution

1. Step 1: Understand symmetry in trees

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

2. Step 2: Compare with other options

   Having the same number of nodes, preorder traversal, or height does not guarantee mirror symmetry.

3. Final Answer:

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

4. Quick Check:

   Symmetry means mirror image = A [OK]

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

Common Mistakes:

• Confusing same size with symmetry
• Thinking same traversal means symmetric
• Assuming same height means symmetric

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

easy

A. int isSymmetric(struct Node* root1, struct Node* root2);

B. void isSymmetric(struct Node root1, struct Node root2);

C. int isSymmetric(struct Node root1, struct Node root2);

D. int isSymmetric(Node* root1, Node root2);

Solution

1. Step 1: Understand pointer usage in C for trees

   Tree nodes are accessed via pointers, so parameters must be pointers to struct Node.

2. Step 2: Analyze each option

   int isSymmetric(struct Node* root1, struct Node* root2); uses pointers correctly and returns int for boolean result. Options A and B pass structs by value, which is inefficient and incorrect. int isSymmetric(Node* root1, Node root2); mixes pointer and non-pointer types incorrectly.

3. Final Answer:

   int isSymmetric(struct Node* root1, struct Node* root2); -> Option A

4. Quick Check:

   Use pointers for tree nodes in C function [OK]

Hint: Use pointers to struct Node for tree parameters [OK]

Common Mistakes:

• Passing structs by value instead of pointers
• Mixing pointer and non-pointer parameters
• Using void return type instead of int for boolean

3. What will be the output of the following code snippet checking symmetry of two trees?

struct Node {
  int val;
  struct Node* left;
  struct Node* right;
};

int isMirror(struct Node* t1, struct Node* t2) {
  if (!t1 && !t2) return 1;
  if (!t1 || !t2) return 0;
  return (t1->val == t2->val) && isMirror(t1->left, t2->right) && isMirror(t1->right, t2->left);
}

int main() {
  struct Node n1 = {1, NULL, NULL};
  struct Node n2 = {1, NULL, NULL};
  printf("%d", isMirror(&n1, &n2));
  return 0;
}

medium

A. 0

B. 1

C. Compilation error

D. Runtime error

Solution

1. Step 1: Analyze base cases and node values

   Both nodes have value 1 and no children, so both left and right pointers are NULL.

2. Step 2: Evaluate recursive calls

   isMirror returns 1 if both nodes are NULL (true). Since both left and right children are NULL, recursive calls return 1. Values match, so overall returns 1.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Identical single nodes are symmetric = 1 [OK]

Hint: Single nodes with same value are symmetric [OK]

Common Mistakes:

• Confusing NULL children as asymmetric
• Expecting 0 for identical nodes
• Misreading recursive base cases

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

int isMirror(struct Node* t1, struct Node* t2) {
  if (t1 == NULL && t2 == NULL) return 1;
  if (t1 == NULL || t2 == NULL) return 0;
  if (t1->val != t2->val) return 0;
  return isMirror(t1->left, t2->left) && isMirror(t1->right, t2->right);
}

medium

A. Should use OR instead of AND in return statement

B. Missing check for NULL nodes

C. Return type should be void

D. Should compare t1->left with t2->right and t1->right with t2->left

Solution

1. Step 1: Understand mirror symmetry check

   Mirror symmetry requires left subtree of one tree to match right subtree of the other and vice versa.

2. Step 2: Identify incorrect recursive calls

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

3. Final Answer:

   Should compare t1->left with t2->right and t1->right with t2->left -> Option D

4. Quick Check:

   Mirror means cross compare left and right subtrees [OK]

Hint: Mirror check swaps left and right subtrees [OK]

Common Mistakes:

• Comparing left with left and right with right
• Ignoring NULL checks
• Using wrong logical operators

5. Given two binary trees, how would you implement a function to check if they are symmetric using a queue (iterative approach)? Which step is essential in this approach?

hard

A. Compare only root nodes and ignore children

B. Traverse both trees in preorder and compare arrays

C. Enqueue pairs of nodes to compare left of one with right of other

D. Use a stack to store nodes from only one tree

Solution

1. Step 1: Understand iterative symmetry check

   Use a queue to store pairs of nodes to compare their values and subtrees iteratively.

2. Step 2: Identify correct node pairing

   Pairs must be left child of one tree with right child of the other to check mirror symmetry.

3. Step 3: Reject incorrect approaches

   Preorder traversal comparison or using stack for one tree only does not check mirror symmetry. Ignoring children misses the point.

4. Final Answer:

   Enqueue pairs of nodes to compare left of one with right of other -> Option C

5. Quick Check:

   Queue pairs left-right nodes for iterative mirror check [OK]

Hint: Queue pairs left and right nodes to check symmetry iteratively [OK]

Common Mistakes:

• Comparing same side children instead of opposite
• Using preorder traversal arrays
• Ignoring children nodes in comparison