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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Rust

if p.is_none() && [1] { return true; }

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

Ap.is_some()

Bq.is_none()

Cq.is_some()

Dq.is_empty()

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if the other node is Some instead of None.

Using incorrect method like is_empty() which doesn't exist for Option.

Both nodes must be None to be symmetric at this point.

2fill in blank

medium

Complete the code to check if one node is None and the other is not.

DSA Rust

if p.is_none() || [1] { return false; }

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

Ap.is_some()

Bq.is_none()

Cq.is_some()

Dq.is_empty()

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if the other node is None instead of Some.

Using non-existent methods like is_empty().

If one node is None and the other is Some, trees are not symmetric.

3fill in blank

hard

Fix the error in comparing the values of two nodes.

DSA Rust

if p.as_ref().unwrap().borrow().val != [1].as_ref().unwrap().borrow().val { return false; }

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

Aq

Broot

Cself

Dp

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing p with itself.

Using undefined variables like self or root.

We compare the value of node p with node q to check symmetry.

4fill in blank

hard

Fill both blanks to recursively check left and right children for symmetry.

DSA Rust

return Self::is_symmetric_helper(p.as_ref().unwrap().borrow().[1].clone(), q.as_ref().unwrap().borrow().[2].clone()) &&
       Self::is_symmetric_helper(p.as_ref().unwrap().borrow().right.clone(), q.as_ref().unwrap().borrow().left.clone());

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

Aleft

Bright

Cval

Dclone

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right incorrectly.

Using val or clone in place of child nodes.

We compare p's left with q's right and p's right with q's left recursively.

5fill in blank

hard

Fill all three blanks to complete the public method that starts the symmetry check.

DSA Rust

pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    if root.is_none() {
        [1] true;
    }
    let root_ref = root.as_ref().unwrap();
    Self::is_symmetric_helper(root_ref.borrow().[2].clone(), root_ref.borrow().[3].clone())
}

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

Areturn

Bright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Not returning true when root is None.

Swapping left and right children in the helper call.

If root is None, tree is symmetric by definition. Then we compare left and right children.

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