Check if Two Trees are Symmetric in DSA Rust - 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 work increase when the trees have more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn is_symmetric(t1: &Option>, t2: &Option>) -> bool {
    match (t1, t2) {
        (None, None) => true,
        (Some(n1), Some(n2)) => {
            n1.val == n2.val &&
            is_symmetric(&n1.left, &n2.right) &&
            is_symmetric(&n1.right, &n2.left)
        },
        _ => false,
    }
}

This function checks if two trees are mirror images by comparing nodes 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 trees, visiting all nodes.

How Execution Grows With Input

As the number of nodes grows, the function checks each node pair once.

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

Pattern observation: The work grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check symmetry grows linearly with the number of nodes in the trees.

Common Mistake

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

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

Interview Connect

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

Self-Check

"What if we changed the function to check only one subtree instead of both? 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, meaning left subtree of one matches right subtree of the other and vice versa.

2. Step 2: Compare structure and 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 = C [OK]

Hint: Symmetric trees mirror each other exactly [OK]

Common Mistakes:

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

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

easy

A. fn is_symmetric(t1: Option<TreeNode>, t2: Option<TreeNode>) -> bool

B. fn is_symmetric(t1: &Option, t2: &Option) -> bool

C. fn is_symmetric(t1: &TreeNode, t2: &TreeNode) -> Option<bool>

D. fn is_symmetric(t1: TreeNode, t2: TreeNode) -> bool

Solution

1. Step 1: Understand Rust tree node representation

   Binary trees in Rust are often represented as Option to handle empty nodes safely.

2. Step 2: Choose correct signature

   Passing references to Option avoids ownership issues and returns bool for symmetric check.

3. Final Answer:

   fn is_symmetric(t1: &Option, t2: &Option) -> bool -> Option B

4. Quick Check:

   Use references to Option and bool return [OK]

Hint: Use references to Option for tree nodes [OK]

Common Mistakes:

• Passing TreeNode by value causing ownership errors
• Returning Option<bool> instead of bool
• Using non-Option types for nullable nodes

3. Given the following Rust code snippet, what will be the output?

struct TreeNode {
    val: i32,
    left: Option,
    right: Option,
}

fn is_mirror(t1: &Option, t2: &Option) -> bool {
    match (t1, t2) {
        (None, None) => true,
        (Some(n1), Some(n2)) => {
            n1.val == n2.val &&
            is_mirror(&n1.left, &n2.right) &&
            is_mirror(&n1.right, &n2.left)
        },
        _ => false,
    }
}

fn main() {
    let tree1 = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode { val: 2, left: None, right: None })),
        right: Some(Box::new(TreeNode { val: 3, left: None, right: None })),
    }));
    let tree2 = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode { val: 3, left: None, right: None })),
        right: Some(Box::new(TreeNode { val: 2, left: None, right: None })),
    }));
    println!("{}", is_mirror(&tree1, &tree2));
}

medium

A. Runtime panic

B. false

C. Compilation error

D. true

Solution

1. Step 1: Analyze tree structure

   tree1 has root 1, left child 2, right child 3; tree2 has root 1, left child 3, right child 2.

2. Step 2: Check mirror condition

   Left subtree of tree1 (2) matches right subtree of tree2 (2), and right subtree of tree1 (3) matches left subtree of tree2 (3), values match.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Mirror check returns true [OK]

Hint: Mirror values and subtree positions must match [OK]

Common Mistakes:

• Confusing left and right subtree positions
• Assuming same structure means symmetric
• Ignoring node values in comparison

4. Identify the bug in this Rust function that checks if two trees are symmetric:

fn is_mirror(t1: &Option, t2: &Option) -> bool {
    if t1.is_none() && t2.is_none() {
        return true;
    }
    if t1.is_none() || t2.is_none() {
        return false;
    }
    let n1 = t1.as_ref().unwrap();
    let n2 = t2.as_ref().unwrap();
    n1.val == n2.val &&
    is_mirror(&n1.left, &n2.left) &&
    is_mirror(&n1.right, &n2.right)
}

medium

A. It compares left subtree with left subtree instead of left with right.

B. It unwraps Option without checking for None.

C. It returns false when both trees are empty.

D. It uses incorrect return type.

Solution

1. Step 1: Review subtree comparisons

   The function compares n1.left with n2.left and n1.right with n2.right, which checks for equality, not mirror symmetry.

2. Step 2: Identify correct mirror check

   For symmetry, n1.left should be compared with n2.right and n1.right with n2.left.

3. Final Answer:

   It compares left subtree with left subtree instead of left with right. -> Option A

4. Quick Check:

   Mirror check swaps left and right subtrees [OK]

Hint: Mirror check swaps left and right subtrees [OK]

Common Mistakes:

• Comparing left with left and right with right
• Forgetting to check both subtrees recursively
• Ignoring None cases

5. You want to check if a single binary tree is symmetric around its center. Which approach correctly uses the two-tree symmetric check function is_mirror to solve this?

hard

A. Call is_mirror(&root, &root) to compare the tree with itself.

B. Call is_mirror(&root.left, &root.left) to compare left subtree with itself.

C. Call is_mirror(&root.left, &root.right) to compare left and right subtrees.

D. Call is_mirror(&root.right, &root.right) to compare right subtree with itself.

Solution

1. Step 1: Understand single tree symmetry

   A tree is symmetric if its left subtree is a mirror of its right subtree.

2. Step 2: Use is_mirror function correctly

   Comparing root.left and root.right subtrees with is_mirror checks this mirror condition.

3. Final Answer:

   Call is_mirror(&root.left, &root.right) to compare left and right subtrees. -> Option C

4. Quick Check:

   Compare left and right subtrees for symmetry [OK]

Hint: Check left and right subtrees for mirror symmetry [OK]

Common Mistakes:

• Comparing tree with itself instead of subtrees
• Comparing left subtree with itself
• Comparing right subtree with itself