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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if two nodes are both None.

DSA Python

def is_symmetric_helper(node1, node2):
    if node1 is [1] and node2 is None:
        return True

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

AFalse

Bnot None

CNone

DTrue

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of 'is' to check for None.

Checking only one node for None.

We check if both nodes are None to confirm symmetry at this branch.

2fill in blank

medium

Complete the code to check if the values of two nodes are equal.

DSA Python

def is_symmetric_helper(node1, node2):
    if node1 is None or node2 is None:
        return False
    if node1.val [1] node2.val:
        return False

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

A!=

B==

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' causing wrong logic.

Using greater or less than operators incorrectly.

If the values are not equal, the trees are not symmetric.

3fill in blank

hard

Fix the error in the recursive call to check symmetry of subtrees.

DSA Python

def is_symmetric_helper(node1, node2):
    if node1 is None and node2 is None:
        return True
    if node1 is None or node2 is None:
        return False
    if node1.val != node2.val:
        return False
    return is_symmetric_helper(node1.left, [1]) and is_symmetric_helper(node1.right, node2.left)

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

Anode2.right

Bnode2.left

Cnode1.right

Dnode1.left

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing left subtree with left subtree instead of right.

Mixing up node1 and node2 in recursive calls.

We compare node1's left subtree with node2's right subtree for symmetry.

4fill in blank

hard

Fill both blanks to complete the main function that starts the symmetry check.

DSA Python

def is_symmetric(root):
    if root is None:
        return True
    return is_symmetric_helper(root.[1], root.[2])

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

Aleft

Bright

Cval

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'val' or 'node' instead of child nodes.

Comparing root with itself.

The symmetry check starts by comparing the left and right children of the root.

5fill in blank

hard

Fill all three blanks to complete the recursive symmetry check combining conditions.

DSA Python

def is_symmetric_helper(node1, node2):
    if node1 is None and node2 is None:
        return True
    if node1 is None or node2 is None:
        return False
    if node1.val != node2.val:
        return False
    return is_symmetric_helper(node1.[1], node2.[2]) and is_symmetric_helper(node1.[3], node2.[4])

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

Aleft

Bright

Cval

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same child for both recursive calls.

Mixing up left and right in recursive calls.

The function compares node1's left with node2's right and node1's right with node2's left recursively.

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