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, 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

How do you compare nodes when checking for symmetry?

Compare the values of the current nodes. Then recursively check if the left child of the first tree is symmetric with the right child of the second tree, and the right child of the first tree is symmetric with the left child of the second tree.

intermediate

Why is recursion a good approach to check tree symmetry?

Because symmetry involves comparing subtrees in a mirrored way, recursion naturally handles this by breaking the problem into smaller symmetric checks on child nodes.

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 you return if both input trees are null when checking symmetry?

A True

B False

C Null

D Error

If both trees are empty, they are symmetric by definition.

When comparing two nodes for symmetry, which children do you compare?

A Left child of first tree with left child of second tree

B Right child of first tree with right child of second tree

C Left child of first tree with right child of second tree

D None of the above

Symmetry means left subtree of one matches right subtree of the other.

If the root values of two trees differ, are the trees symmetric?

A Yes

B No

C Only if subtrees are symmetric

D Depends on tree height

Root values must be equal for trees to be symmetric.

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

A Recursion

B Iteration

C Sorting

D Hashing

Recursion is used to compare mirrored subtrees.

What is the worst-case time complexity to check if two trees with n nodes are symmetric?

A O(n^2)

B O(log n)

C O(1)

D O(n)

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

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

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

Describe the time complexity and why it is that way when checking two trees for symmetry.

Consider how many nodes you check and how often.

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