Check if Two Trees are Symmetric in DSA C - 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.

bool isMirror(struct TreeNode* t1, struct TreeNode* t2) {
    if (t1 == NULL && t2 == NULL) return true;
    if (t1 == NULL || t2 == NULL) return false;
    return (t1->val == t2->val) 
        && isMirror(t1->left, t2->right) 
        && isMirror(t1->right, t2->left);
}

bool isSymmetric(struct TreeNode* root) {
    return isMirror(root, root);
}
    

This code checks if a tree is symmetric by comparing its left and right sides recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls comparing pairs of nodes from left and right subtrees.
• How many times: Once for each node in the tree, visiting all nodes to check symmetry.

How Execution Grows With Input

As the tree grows, the function checks each node once by comparing pairs of nodes.

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

Pattern observation: The number of operations 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 tree, 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 the full tree size.

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 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 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