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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if two nodes are both null.

DSA C++

if (root1 == nullptr && root2 [1] nullptr) return true;

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

A==

B&&

C!=

D||

Attempts:

3 left

💡 Hint

Common Mistakes

Using && causes a syntax error (root2 && nullptr).

Using != checks if root2 is not null, which is incorrect.

Using || changes the logic to OR, which is wrong.

We check if both root1 and root2 are null by using == to compare root2 with nullptr.

2fill in blank

medium

Complete the code to check if one node is null and the other is not.

DSA C++

if (root1 == nullptr || root2 [1] nullptr) return false;

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

A&&

B==

C!=

D||

Attempts:

3 left

💡 Hint

Common Mistakes

Using != fails to detect root1 non-null and root2 null.

Using && requires both null, already handled.

Using || is misplaced; it's already between conditions.

Since the both-null case is handled earlier, we check if root1 == nullptr || root2 == nullptr to catch exactly one null.

3fill in blank

hard

Fix the error in comparing node values for symmetry.

DSA C++

if (root1->val [1] root2->val) return false;

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

A==

B>

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using == returns false when values are equal, which is wrong.

Using < or > is not correct for equality check.

We return false if the values are not equal, so we use !=.

4fill in blank

hard

Fill both blanks to recursively check left and right subtrees for symmetry.

DSA C++

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

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

Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right incorrectly.

Using 'val' or 'parent' instead of child pointers.

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 calling the helper.

DSA C++

bool isSymmetric(TreeNode* root) {
    if (root == nullptr) return true;
    return isSymmetricHelper(root->[1], root->[2]);
}

bool isSymmetricHelper(TreeNode* root1, TreeNode* root2) {
    if (root1 == nullptr && root2 == nullptr) return true;
    if (root1 == nullptr || root2 [3] nullptr) return false;
    if (root1->val != root2->val) return false;
    return isSymmetricHelper(root1->left, root2->right) && isSymmetricHelper(root1->right, root2->left);
}

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

Aleft

Bright

C!=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Passing root itself instead of children.

Using != instead of == in null check.

The main function calls helper with root's left and right children. The helper checks if one node is null and the other is not using ==.

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 = A [OK]

Hint: Symmetric trees mirror each other perfectly [OK]

Common Mistakes:

• Thinking same number of nodes means symmetric
• Only comparing root values
• Ignoring structure when checking symmetry

2. Which of the following C++ function signatures correctly represents a function to check if two trees are symmetric?

easy

A. int isSymmetric(TreeNode* root1, TreeNode* root2);

B. bool isSymmetric(TreeNode* root1, TreeNode* root2);

C. void isSymmetric(TreeNode root1, TreeNode root2);

D. bool isSymmetric(TreeNode root1, TreeNode root2);

Solution

1. Step 1: Identify correct return type

   The function should return a boolean indicating symmetry, so return type is bool.

2. Step 2: Use pointers for tree nodes

   Tree nodes are usually passed as pointers to avoid copying and to handle null nodes.

3. Final Answer:

   bool isSymmetric(TreeNode* root1, TreeNode* root2); -> Option B

4. Quick Check:

   Return bool and use pointers = C [OK]

Hint: Use bool return and pointers for tree nodes [OK]

Common Mistakes:

• Using void return type
• Passing nodes by value instead of pointer
• Using int instead of bool for result

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

#include <iostream>

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

bool isMirror(TreeNode* t1, TreeNode* 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);
}

int main() {
  TreeNode* root1 = new TreeNode(1);
  root1->left = new TreeNode(2);
  root1->right = new TreeNode(3);

  TreeNode* root2 = new TreeNode(1);
  root2->left = new TreeNode(3);
  root2->right = new TreeNode(2);

  if (isMirror(root1, root2))
    std::cout << "Symmetric";
  else
    std::cout << "Not Symmetric";
  return 0;
}

medium

A. Symmetric

B. Not Symmetric

C. Compilation Error

D. Runtime Error

Solution

1. Step 1: Analyze tree structures

   root1 has left=2, right=3; root2 has left=3, right=2, which are mirror images.

2. Step 2: Trace isMirror function

   Values match and recursive calls check left-right and right-left subtrees, confirming symmetry.

3. Final Answer:

   Symmetric -> Option A

4. Quick Check:

   Mirror check returns true = Symmetric [OK]

Hint: Check mirrored node values and subtree positions [OK]

Common Mistakes:

• Confusing left and right subtree checks
• Ignoring null node base cases
• Assuming same structure means symmetric

4. Identify the bug in the following code that checks if two trees are symmetric:

bool isMirror(TreeNode* t1, TreeNode* 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 uses incorrect return type.

B. It does not check for null nodes.

C. It returns true when both nodes are null.

D. It compares left subtree with left subtree instead of left with right.

Solution

1. Step 1: Review recursive calls for symmetry

   Symmetry requires comparing left subtree of one tree with right subtree of the other.

2. Step 2: Identify incorrect subtree comparisons

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

3. Final Answer:

   It compares left subtree with left subtree instead of left with right. -> Option D

4. Quick Check:

   Mirror needs left-right swap, code misses it = D [OK]

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

Common Mistakes:

• Comparing same side subtrees instead of mirrored sides
• Ignoring null checks
• Confusing identical and symmetric trees

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

hard

A. Use a recursive function that compares left subtree of first tree with right subtree of second tree and vice versa.

B. Traverse both trees in preorder and compare node values in order.

C. Check if both trees have the same number of nodes and same root value.

D. Use a queue to perform level order traversal and compare nodes at each level without mirroring.

Solution

1. Step 1: Understand symmetry check logic

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

2. Step 2: Evaluate other approaches

   Preorder traversal or node count alone do not guarantee symmetry; level order without mirroring misses mirror structure.

3. Final Answer:

   Use a recursive function that compares left subtree of first tree with right subtree of second tree and vice versa. -> Option A

4. Quick Check:

   Recursive mirror comparison solves symmetry = A [OK]

Hint: Recursively compare left with right subtree for symmetry [OK]

Common Mistakes:

• Using simple traversal without mirroring
• Comparing only node counts or root values
• Ignoring subtree structure in mirror form