Check if Two Trees are Symmetric in DSA Python - 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.

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

📋 What You'll Learn

Create two binary trees using the TreeNode class with exact node values

Create a helper function called is_mirror that takes two nodes and checks if they are mirror images

Create a function called are_symmetric that uses is_mirror to check if two trees are symmetric

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 comparison, and validating mirrored data structures.

💼 Career

Understanding tree symmetry helps in software development roles involving data structures, algorithms, and system design.

1

Create two binary trees

Create a class called TreeNode with an __init__ method that takes val, left, and right. Then create two trees called tree1 and tree2 with these exact structures:

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

DSA Python

# Define TreeNode class and create tree1 and tree2
# Your code here

Hint

Define the TreeNode class first. Then create tree1 with root value 1, left child 2, right child 3. Similarly, create tree2 with root 1, left child 3, right child 2.

2

Create the helper function is_mirror

Create a function called is_mirror that takes two parameters node1 and node2. It should return True if both nodes are None, return False if only one is None, and otherwise check if node1.val == node2.val and recursively check is_mirror(node1.left, node2.right) and is_mirror(node1.right, node2.left).

DSA Python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

tree1 = TreeNode(1, TreeNode(2), TreeNode(3))
tree2 = TreeNode(1, TreeNode(3), TreeNode(2))

# Define the is_mirror function below
# Your code here

Hint

Check if both nodes are None, then return True. If only one is None, return False. Otherwise, check if values are equal and recursively check left and right children in mirror order.

3

Create the function are_symmetric

Create a function called are_symmetric that takes two parameters root1 and root2. It should return the result of calling is_mirror(root1, root2).

DSA Python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

tree1 = TreeNode(1, TreeNode(2), TreeNode(3))
tree2 = TreeNode(1, TreeNode(3), TreeNode(2))

def is_mirror(node1, node2):
    if node1 is None and node2 is None:
        return True
    if node1 is None or node2 is None:
        return False
    return (node1.val == node2.val and
            is_mirror(node1.left, node2.right) and
            is_mirror(node1.right, node2.left))

# Define are_symmetric function below
# Your code here

Hint

Simply return the result of is_mirror(root1, root2).

4

Print if the two trees are symmetric

Use print to display the result of calling are_symmetric(tree1, tree2).

DSA Python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

tree1 = TreeNode(1, TreeNode(2), TreeNode(3))
tree2 = TreeNode(1, TreeNode(3), TreeNode(2))

def is_mirror(node1, node2):
    if node1 is None and node2 is None:
        return True
    if node1 is None or node2 is None:
        return False
    return (node1.val == node2.val and
            is_mirror(node1.left, node2.right) and
            is_mirror(node1.right, node2.left))

def are_symmetric(root1, root2):
    return is_mirror(root1, root2)

# Print the result of are_symmetric(tree1, tree2) below
# Your code here

Hint

Call print(are_symmetric(tree1, tree2)) to show if the trees are symmetric.

Practice

1. What does it mean for two binary trees to be symmetric?

easy

A. They are mirror images of each other around their center.

B. They have the same number of nodes.

C. They have the same root value only.

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 A

4. Quick Check:

   Symmetry means mirror image = A [OK]

Hint: Symmetric trees mirror each other exactly [OK]

Common Mistakes:

• Thinking same number of nodes means symmetric
• Ignoring node values in comparison
• Checking only left or right subtree

2. Which of the following Python function signatures is correct to check if two trees are symmetric?

easy

A. def is_symmetric(tree1, tree2)

B. def is_symmetric(tree1 tree2):

C. def is_symmetric(tree1, tree2) -> bool:

D. def is_symmetric(tree1, tree2): return True

Solution

1. Step 1: Check syntax for function definition

   Python functions require parentheses around parameters and a colon at the end.

2. Step 2: Identify correct type hint usage

   def is_symmetric(tree1, tree2) -> bool: uses correct syntax with type hint and colon, making it valid and clear.

3. Final Answer:

   def is_symmetric(tree1, tree2) -> bool: -> Option C

4. Quick Check:

   Correct Python function syntax with type hint = D [OK]

Hint: Use parentheses and colon; type hints are optional but clear [OK]

Common Mistakes:

• Missing comma between parameters
• Omitting colon at function end
• Incorrect parameter syntax without parentheses

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

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def is_mirror(t1, t2):
    if not t1 and not t2:
        return True
    if not t1 or not t2:
        return False
    return (t1.val == t2.val) and is_mirror(t1.left, t2.right) and is_mirror(t1.right, t2.left)

root1 = Node(1)
root1.left = Node(2)
root1.right = Node(3)

root2 = Node(1)
root2.left = Node(3)
root2.right = Node(2)

print(is_mirror(root1, root2))

medium

A. True

B. False

C. None

D. Error

Solution

1. Step 1: Check base cases for empty nodes

   Both trees have root 1, and their left and right children are swapped but equal in value.

2. Step 2: Recursively verify mirror condition

   Left of root1 (2) matches right of root2 (2), and right of root1 (3) matches left of root2 (3), so all conditions return True.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Mirror check returns True for swapped children = A [OK]

Hint: Mirror means left of one equals right of other [OK]

Common Mistakes:

• Confusing left and right children in recursion
• Ignoring base cases for None nodes
• Assuming same structure means symmetric

4. Identify the error in this code snippet that checks if two trees are symmetric:

def is_symmetric(t1, t2):
    if t1 is None and t2 is None:
        return True
    if t1 is None or t2 is None:
        return False
    if t1.val != t2.val:
        return False
    return is_symmetric(t1.left, t2.left) and is_symmetric(t1.right, t2.right)

medium

A. It does not check for None nodes.

B. It compares left with left and right with right instead of mirror sides.

C. It returns True too early without checking values.

D. It uses 'is' instead of '==' for value comparison.

Solution

1. Step 1: Analyze recursive calls for symmetry

   The function compares left subtree of t1 with left subtree of t2 and right with right, which is incorrect for mirror symmetry.

2. Step 2: Correct mirror comparison

   It should compare left of t1 with right of t2 and right of t1 with left of t2 to check mirror symmetry properly.

3. Final Answer:

   It compares left with left and right with right instead of mirror sides. -> Option B

4. Quick Check:

   Mirror check requires cross comparison = B [OK]

Hint: Mirror means cross compare left with right [OK]

Common Mistakes:

• Comparing same sides instead of mirror sides
• Confusing 'is' and '==' for values
• Missing base cases for None nodes

5. Given two binary trees, write a function to check if they are symmetric considering that one tree might be None. Which approach correctly handles this edge case?

hard

A. Return False immediately if either tree is None.

B. Ignore None checks and compare node values only.

C. Only compare root values and ignore subtrees.

D. Return True if both trees are None, else check mirror recursively.

Solution

1. Step 1: Handle None edge cases carefully

   If both trees are None, they are symmetric by definition, so return True.

2. Step 2: Recursively check mirror symmetry

   If only one is None, return False. Otherwise, compare root values and recursively check left of one with right of other and vice versa.

3. Final Answer:

   Return True if both trees are None, else check mirror recursively. -> Option D

4. Quick Check:

   Both None means symmetric, else recurse mirror check = C [OK]

Hint: Both None means symmetric; one None means not symmetric [OK]

Common Mistakes:

• Returning False if either tree is None without checking both
• Skipping recursive mirror checks
• Comparing only root values