Mirror a Binary Tree in DSA Python - Time & Space Complexity

Time Complexity: Mirror a Binary Tree

O(n)

Understanding Time Complexity

We want to know how long it takes to flip a binary tree so that left and right children swap places.

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def mirror_tree(node):
    if node is None:
        return
    node.left, node.right = node.right, node.left
    mirror_tree(node.left)
    mirror_tree(node.right)

This code swaps left and right children of every node in the tree recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once to swap its children.
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

Each node is visited once, so the work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 swaps
100	About 100 swaps
1000	About 1000 swaps

Pattern observation: Doubling the nodes roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to mirror the tree grows linearly with the number of nodes.

Common Mistake

[X] Wrong: "The time depends on the height of the tree only."

[OK] Correct: Even if the tree is tall or short, every node must be visited once, so total time depends on total nodes, not just height.

Interview Connect

Understanding this helps you reason about recursive tree problems and shows you can analyze how work scales with input size.

Self-Check

"What if we changed the code to use an iterative approach with a stack? How would the time complexity change?"

Practice

1. What does mirroring a binary tree mean?
Mirror a Binary Tree means:

easy

A. Swapping the left and right children of every node

B. Deleting all leaf nodes

C. Reversing the values of nodes in each level

D. Sorting the nodes in ascending order

Solution

1. Step 1: Understand the mirror operation

   Mirroring a binary tree means swapping the left and right children of every node in the tree.

2. Step 2: Compare with other options

   Deleting leaves, reversing values, or sorting nodes are different operations and do not create a mirror image.

3. Final Answer:

   Swapping the left and right children of every node -> Option A

4. Quick Check:

   Mirror = Swap children [OK]

Hint: Mirror means swap left and right children at all nodes [OK]

Common Mistakes:

• Thinking mirror means reversing node values only
• Confusing mirror with tree traversal
• Assuming mirror deletes nodes

2. Which of the following is the correct Python function header to define a recursive mirror function for a binary tree node named root?

easy

A. def mirror(root):

B. def mirror(root) -> int:

C. def mirror(root) -> None:

D. def mirror(root) -> str:

Solution

1. Step 1: Identify function purpose and return type

   The mirror function modifies the tree in place and returns nothing, so return type is None.

2. Step 2: Check syntax correctness

   def mirror(root) -> None: uses correct Python syntax with type hint for no return value.

3. Final Answer:

   def mirror(root) -> None: -> Option C

4. Quick Check:

   Function returns nothing = None [OK]

Hint: Mirror modifies tree, returns nothing, so use -> None [OK]

Common Mistakes:

• Using wrong return type like int or str
• Omitting parentheses in function definition
• Missing colon at end of function header

3. What will be the output of the following code after calling mirror(root)? Assume the tree is:
1
/ \
2 3

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

def mirror(root) -> None:
    if root is None:
        return
    root.left, root.right = root.right, root.left
    mirror(root.left)
    mirror(root.right)

root = Node(1)
root.left = Node(2)
root.right = Node(3)
mirror(root)
print(root.left.val, root.right.val)

medium

A. 1 3

B. 3 2

C. 2 3

D. 3 1

Solution

1. Step 1: Understand initial tree structure

   Root is 1, left child 2, right child 3.

2. Step 2: Apply mirror function

   Mirror swaps left and right children at root, so left becomes 3 and right becomes 2.

3. Final Answer:

   3 2 -> Option B

4. Quick Check:

   Swapped children = 3 2 [OK]

Hint: Mirror swaps children, so left=3 and right=2 after mirror [OK]

Common Mistakes:

• Not swapping children correctly
• Printing values before mirror call
• Confusing left and right child values

4. Find the error in this mirror function code:

def mirror(root):
    if root is None:
        return
    mirror(root.left)
    mirror(root.right)
    root.left = root.right
    root.right = root.left

medium

A. Swapping left and right children is done incorrectly

B. Base case is missing

C. Function does not call itself recursively

D. No return statement in function

Solution

1. Step 1: Analyze swapping logic

   Assigning root.left = root.right then root.right = root.left sets both to same child, losing original left.

2. Step 2: Correct swapping method

   Use simultaneous swap: root.left, root.right = root.right, root.left to avoid overwriting.

3. Final Answer:

   Swapping left and right children is done incorrectly -> Option A

4. Quick Check:

   Swap must be simultaneous to avoid overwrite [OK]

Hint: Swap children simultaneously to avoid overwriting [OK]

Common Mistakes:

• Swapping children one by one causing data loss
• Forgetting base case for recursion
• Not swapping children at all

5. Given a binary tree, write a function to check if the tree is symmetric (a mirror of itself). Which approach correctly uses the mirror concept?

hard

A. Traverse tree in preorder and compare with inorder traversal

B. Check if left and right subtrees have the same number of nodes

C. Check if all node values are equal in left and right subtrees

D. Compare left subtree with right subtree by recursively checking if left.left matches right.right and left.right matches right.left

Solution

1. Step 1: Understand symmetry in binary trees

   A symmetric tree is a mirror of itself, so left subtree must mirror right subtree.

2. Step 2: Use recursive comparison

   Check if left.left matches right.right and left.right matches right.left recursively to confirm mirror symmetry.

3. Final Answer:

   Compare left subtree with right subtree by recursively checking if left.left matches right.right and left.right matches right.left -> Option D

4. Quick Check:

   Symmetry = recursive mirror check [OK]

Hint: Symmetry means left subtree mirrors right subtree recursively [OK]

Common Mistakes:

• Only comparing node counts, not structure
• Comparing node values without structure
• Using traversal orders that don't check mirror