Check if Two Trees are Symmetric in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if both trees are NULL, which means they are symmetric.

DSA C

if (root1 == NULL && root2 == [1]) {
    return 1;
}

Drag options to blanks, or click blank then click option'

Aroot1

B0

Croot2

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Checking only one tree for NULL

Using 0 instead of NULL

Both trees being NULL means they are symmetric, so we check if root2 is NULL.

2fill in blank

medium

Complete the code to check if one tree is NULL and the other is not, which means they are not symmetric.

DSA C

if (root1 == NULL || root2 == [1]) {
    return 0;
}

Drag options to blanks, or click blank then click option'

ANULL

Broot1

Croot2

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using AND instead of OR

Returning 1 instead of 0

If one tree is NULL and the other is not, they cannot be symmetric.

3fill in blank

hard

Fix the error in the comparison of node values to check if they are equal.

DSA C

if (root1->val [1] root2->val) {
    return 0;
}

Drag options to blanks, or click blank then click option'

A!=

B==

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using == instead of !=

Using > or < which only checks order

If the values are not equal, the trees are not symmetric, so we return 0.

4fill in blank

hard

Fill both blanks to recursively check if left subtree of root1 is symmetric with right subtree of root2 and vice versa.

DSA C

return isSymmetricHelper(root1->[1], root2->[2]) && isSymmetricHelper(root1->right, root2->left);

Drag options to blanks, or click blank then click option'

Aleft

Bright

Croot

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using left for both subtrees

Using right for both subtrees

We compare root1's left with root2's right and root1's right with root2's left for symmetry.

5fill in blank

hard

Fill all three blanks to complete the main function that calls the helper to check if a tree is symmetric with itself.

DSA C

int isSymmetric(struct TreeNode* root) {
    if (root == NULL) return 1;
    return isSymmetricHelper(root, [1]);
}

Drag options to blanks, or click blank then click option'

Aroot->left

Broot->right

Croot

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Passing root->left or root->right instead of root

Passing NULL as second argument

We check if the tree is symmetric by comparing it with itself using the helper function.

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