Check if Two Trees are Symmetric in DSA Go - Time & Space Complexity

Time Complexity: Check if Two Trees are Symmetric

O(n)

Understanding Time Complexity

We want to know how the time needed to check if two trees are mirror images grows as the trees get bigger.

How does the number of steps change when the trees have more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func isSymmetric(t1, t2 *TreeNode) bool {
    if t1 == nil && t2 == nil {
        return true
    }
    if t1 == nil || t2 == nil {
        return false
    }
    return t1.Val == t2.Val && isSymmetric(t1.Left, t2.Right) && isSymmetric(t1.Right, t2.Left)
}

This code checks if two trees are mirror images by comparing nodes and their opposite children recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls comparing pairs of nodes.
• How many times: Once for each node in the tree.

How Execution Grows With Input

Each node pair is checked once, so the steps grow directly with the number of nodes.

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

Pattern observation: The work grows in a straight line as the tree size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to check symmetry grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The function only checks half the nodes, so it runs in half the time."

[OK] Correct: Even though it compares nodes in pairs, every node is visited once, so the time still grows with all nodes.

Interview Connect

Understanding how recursive tree checks scale helps you explain your approach clearly and confidently in interviews.

Self-Check

"What if we changed the code to check only one subtree against itself instead of two? How would the time complexity change?"

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 [OK]

Hint: Symmetric trees mirror each other exactly [OK]

Common Mistakes:

• Thinking same number of nodes means symmetric
• Only comparing root values
• Ignoring structure differences

2. Which Go function signature correctly represents a function to check if two trees are symmetric?

easy

A. func isSymmetric(t1 *TreeNode) bool

B. func isSymmetric(t1 TreeNode, t2 TreeNode) int

C. func isSymmetric(t1 *TreeNode, t2 *TreeNode) bool

D. func isSymmetric(t1 *TreeNode, t2 *TreeNode) string

Solution

1. Step 1: Identify correct parameter types

   Function should take pointers to TreeNode structs for both trees to compare.

2. Step 2: Identify correct return type

   Return type should be bool to indicate symmetric (true) or not (false).

3. Final Answer:

   func isSymmetric(t1 *TreeNode, t2 *TreeNode) bool -> Option C

4. Quick Check:

   Pointer params and bool return [OK]

Hint: Use pointers and bool return for tree comparison [OK]

Common Mistakes:

• Using value instead of pointer for trees
• Returning int or string instead of bool
• Missing one tree parameter

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

package main
import "fmt"

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

func isSymmetric(t1, t2 *TreeNode) bool {
    if t1 == nil && t2 == nil {
        return true
    }
    if t1 == nil || t2 == nil {
        return false
    }
    return t1.Val == t2.Val && isSymmetric(t1.Left, t2.Right) && isSymmetric(t1.Right, t2.Left)
}

func main() {
    root1 := &TreeNode{Val: 1, Left: &TreeNode{Val: 2}, Right: &TreeNode{Val: 2}}
    root2 := &TreeNode{Val: 1, Left: &TreeNode{Val: 2}, Right: &TreeNode{Val: 3}}
    fmt.Println(isSymmetric(root1, root2))
}

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Compare root nodes

   Both roots have value 1, so check children.

2. Step 2: Compare left and right subtrees

   Left subtree of root1 has value 2, right subtree of root2 has value 3, which differ.

3. Final Answer:

   false -> Option B

4. Quick Check:

   Different child values mean not symmetric [OK]

Hint: Check mirrored child values for symmetry [OK]

Common Mistakes:

• Assuming same root means symmetric
• Ignoring mismatch in child nodes
• Confusing symmetric with identical

4. Identify the bug in the following Go function that checks if two trees are symmetric:

func isSymmetric(t1, t2 *TreeNode) bool {
    if t1 == nil && t2 == nil {
        return true
    }
    if t1 == nil || t2 == nil {
        return false
    }
    return t1.Val == t2.Val && isSymmetric(t1.Left, t2.Left) && isSymmetric(t1.Right, t2.Right)
}

medium

A. The function does not check root node values.

B. The base case for nil nodes is missing.

C. The function should return int, not bool.

D. The recursive calls compare wrong child nodes for symmetry.

Solution

1. Step 1: Check recursive calls for symmetry

   Symmetry requires comparing left of one tree with right of the other, but code compares left with left and right with right.

2. Step 2: Identify correct recursive calls

   Correct calls should be isSymmetric(t1.Left, t2.Right) and isSymmetric(t1.Right, t2.Left).

3. Final Answer:

   The recursive calls compare wrong child nodes for symmetry. -> Option D

4. Quick Check:

   Mirror children must be compared [OK]

Hint: Compare left with right child for symmetry [OK]

Common Mistakes:

• Comparing left with left and right with right
• Missing nil base cases
• Returning wrong type

5. Given two binary trees, write a Go function to check if they are symmetric. Which approach correctly handles the edge case where one tree is nil and the other is not?

hard

A. Return false immediately if one tree is nil and the other is not.

B. Return true if one tree is nil and the other is not.

C. Ignore nil checks and compare node values only.

D. Return true if both trees have the same number of nodes.

Solution

1. Step 1: Handle nil tree edge cases

   If one tree is nil and the other is not, they cannot be symmetric, so return false immediately.

2. Step 2: Ensure correct base cases in recursion

   Only return true if both trees are nil at the same time, otherwise false.

3. Final Answer:

   Return false immediately if one tree is nil and the other is not. -> Option A

4. Quick Check:

   Nil mismatch means not symmetric [OK]

Hint: Check nil mismatch first to avoid errors [OK]

Common Mistakes:

• Returning true when one tree is nil
• Skipping nil checks causing runtime errors
• Comparing only node counts