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

Time Complexity: Mirror a Binary Tree

O(n)

Understanding Time Complexity

We want to understand 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.

func mirrorTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    left := mirrorTree(root.Left)
    right := mirrorTree(root.Right)
    root.Left = right
    root.Right = left
    return root
}
    

This code flips the tree by swapping left and right children for every node recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

Each node is visited once to swap its children, 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 complexity is O(log n) because the tree height is log n."

[OK] Correct: Even though the tree height is log n, every node must be visited once, so the total work depends on all nodes, not just the height.

Interview Connect

Understanding this helps you explain how recursive tree algorithms work and how their time depends on the number of nodes, a key skill in many coding interviews.

Self-Check

"What if we changed the code to only mirror the left subtree and not the right? How would the time complexity change?"

Practice

1. What does it mean to mirror a binary tree?

easy

A. Swap the left and right children of every node in the tree

B. Remove all leaf nodes from the tree

C. Convert the tree into a linked list

D. Sort 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.

2. Step 2: Apply the definition to the tree

   By swapping children at every node, the tree becomes a mirror image of the original.

3. Final Answer:

   Swap the left and right children of every node in the tree -> Option A

4. Quick Check:

   Mirror = Swap children [OK]

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

Common Mistakes:

• Thinking mirror means deleting nodes
• Confusing mirror with sorting
• Assuming mirror means flattening the tree

2. Which Go function signature correctly represents a function to mirror a binary tree node?

easy

A. func mirrorTree(root *TreeNode) int

B. func mirrorTree(root TreeNode) TreeNode

C. func mirrorTree(root TreeNode) bool

D. func mirrorTree(root *TreeNode) *TreeNode

Solution

1. Step 1: Understand pointer usage in Go trees

   Binary trees are usually represented with pointers to nodes to allow modification.

2. Step 2: Check return type for mirrored tree

   The function should return a pointer to the root of the mirrored tree.

3. Final Answer:

   func mirrorTree(root *TreeNode) *TreeNode -> Option D

4. Quick Check:

   Pointer input and output for tree nodes [OK]

Hint: Use pointers for tree nodes to modify and return [OK]

Common Mistakes:

• Passing tree nodes by value instead of pointer
• Returning wrong type like int or bool
• Not returning the mirrored tree root

3. What is the output of the following Go code that mirrors a binary tree?

type TreeNode struct {
    Val int
    Left, Right *TreeNode
}

func mirrorTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    left := mirrorTree(root.Left)
    right := mirrorTree(root.Right)
    root.Left, root.Right = right, left
    return root
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    mirrorTree(root)
    fmt.Println(root.Val, root.Left.Val, root.Right.Val)
}

medium

A. 2 1 3

B. 1 2 3

C. 1 3 2

D. 3 1 2

Solution

1. Step 1: Understand the initial tree structure

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

2. Step 2: Apply mirrorTree function

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

3. Final Answer:

   1 3 2 -> Option C

4. Quick Check:

   Swapped children values = 1 3 2 [OK]

Hint: Mirror swaps left and right children recursively [OK]

Common Mistakes:

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

4. Identify the error in this Go function that attempts to mirror a binary tree:

func mirrorTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    mirrorTree(root.Left)
    mirrorTree(root.Right)
    root.Left, root.Right = root.Right, root.Left
    return root
}

medium

A. The base case is missing

B. The recursive calls do not assign the mirrored subtrees back to root.Left and root.Right

C. The function returns the wrong type

D. Swapping children should happen before recursive calls

Solution

1. Step 1: Analyze recursive calls

   The function calls mirrorTree on root.Left and root.Right but ignores their returned mirrored subtrees.

2. Step 2: Check assignment of mirrored subtrees

   Without assigning returned values, the subtrees are not updated, so mirror is incomplete.

3. Final Answer:

   The recursive calls do not assign the mirrored subtrees back to root.Left and root.Right -> Option B

4. Quick Check:

   Assign returned mirrored subtrees to children [OK]

Hint: Always assign recursive results back to node children [OK]

Common Mistakes:

• Ignoring returned subtree pointers
• Swapping children before recursion
• Missing base case (though present here)

5. Given a binary tree, how would you modify the mirror function to also count how many nodes were swapped during the mirroring process?

hard

A. Add a counter variable and increment it each time left and right children are swapped recursively

B. Return the sum of node values instead of mirroring

C. Only swap nodes if their values are even

D. Use a queue to traverse the tree without recursion

Solution

1. Step 1: Understand the goal

   We want to count how many nodes had their children swapped during mirroring.

2. Step 2: Modify mirror function to count swaps

   Each time we swap left and right children at a node, increment a counter and add counts from recursive calls.

3. Final Answer:

   Add a counter variable and increment it each time left and right children are swapped recursively -> Option A

4. Quick Check:

   Count swaps during recursion [OK]

Hint: Count swaps inside recursion when swapping children [OK]

Common Mistakes:

• Counting node values instead of swaps
• Not accumulating counts from recursive calls
• Using traversal without swapping