Check if Two Trees are Symmetric in DSA C++ - Build from Scratch

Check if Two Trees are Symmetric

📖 Scenario: Imagine you have two family trees represented as binary trees. You want to check if these two trees are mirror images of each other, meaning they are symmetric around their center.

🎯 Goal: Build a program that creates two binary trees, sets up a helper function to check symmetry, and then uses this function to determine if the two trees are symmetric.

📋 What You'll Learn

Create two binary trees with exact node values

Create a helper function called isMirror that checks if two trees are mirror images

Create a function called areSymmetric that uses isMirror to check symmetry

Print true if the trees are symmetric, otherwise false

💡 Why This Matters

🌍 Real World

Checking if two trees are symmetric is useful in graphics, data validation, and comparing hierarchical data structures.

💼 Career

Understanding tree symmetry helps in coding interviews and software roles involving tree data structures and recursion.

1

Create Two Binary Trees

Create a struct called Node with an int data, and two pointers left and right. Then create two binary trees called tree1 and tree2 with these exact nodes and structure:

tree1: root 1, left child 2, right child 3
tree2: root 1, left child 3, right child 2

DSA C++

// Define struct Node
// Create tree1 and tree2 with given nodes
// Your code here

Hint

Define the Node struct first. Then create tree1 and tree2 by assigning new nodes with the exact values and structure.

2

Create Helper Function isMirror

Create a function called isMirror that takes two Node* parameters named root1 and root2. This function should return true if the two trees are mirror images of each other, and false otherwise. Use these rules:
- If both nodes are nullptr, return true
- If one is nullptr and the other is not, return false
- Check if root1->data equals root2->data and recursively check root1->left with root2->right and root1->right with root2->left

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* tree1 = new Node(1);
tree1->left = new Node(2);
tree1->right = new Node(3);

Node* tree2 = new Node(1);
tree2->left = new Node(3);
tree2->right = new Node(2);

// Create isMirror function below
// Your code here

Hint

Use recursion to compare the nodes and their children in mirror order.

3

Create Function areSymmetric

Create a function called areSymmetric that takes two Node* parameters named root1 and root2. This function should return the result of calling isMirror(root1, root2).

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* tree1 = new Node(1);
tree1->left = new Node(2);
tree1->right = new Node(3);

Node* tree2 = new Node(1);
tree2->left = new Node(3);
tree2->right = new Node(2);

bool isMirror(Node* root1, Node* root2) {
    if (root1 == nullptr && root2 == nullptr) return true;
    if (root1 == nullptr || root2 == nullptr) return false;
    return (root1->data == root2->data) &&
           isMirror(root1->left, root2->right) &&
           isMirror(root1->right, root2->left);
}

// Create areSymmetric function below
// Your code here

Hint

This function just calls isMirror with the two roots.

4

Print if Trees are Symmetric

Use std::cout to print true if areSymmetric(tree1, tree2) returns true, otherwise print false.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* tree1 = new Node(1);
tree1->left = new Node(2);
tree1->right = new Node(3);

Node* tree2 = new Node(1);
tree2->left = new Node(3);
tree2->right = new Node(2);

bool isMirror(Node* root1, Node* root2) {
    if (root1 == nullptr && root2 == nullptr) return true;
    if (root1 == nullptr || root2 == nullptr) return false;
    return (root1->data == root2->data) &&
           isMirror(root1->left, root2->right) &&
           isMirror(root1->right, root2->left);
}

bool areSymmetric(Node* root1, Node* root2) {
    return isMirror(root1, root2);
}

// Print the result below
// Your code here

Hint

Use std::cout and the ternary operator to print true or false based on the result of areSymmetric(tree1, tree2).

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