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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the root nodes of two trees are both null.

DSA Java

if (root1 == null && root2 [1] null) {
    return true;
}

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

A==

B<

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes wrong logic.

Using comparison operators like '>' or '<' on objects.

We check if both root1 and root2 are null using '=='. If both are null, trees are symmetric at this node.

2fill in blank

medium

Complete the code to check if one of the nodes is null while the other is not.

DSA Java

if (root1 == null || root2 [1] null) {
    return false;
}

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

A!=

B<

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' causes incorrect checks.

Using logical AND instead of OR.

If one node is null and the other is not, trees are not symmetric, so return false.

3fill in blank

hard

Fix the error in comparing the values of the two nodes.

DSA Java

if (root1.val [1] root2.val) {
    return false;
}

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

A==

B>

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' reverses the logic.

Using comparison operators like '>' or '<' instead of equality.

If the values of the two nodes are not equal, the trees are not symmetric.

4fill in blank

hard

Fill both blanks to recursively check symmetry of left and right subtrees.

DSA Java

return isSymmetric(root1.[1], root2.[2]) && isSymmetric(root1.[2], root2.[1]);

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

Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same child (left or right) for both calls.

Using node values instead of child nodes.

We compare root1's left with root2's right and root1's right with root2's left to check mirror symmetry.

5fill in blank

hard

Fill all three blanks to complete the helper method signature and initial call.

DSA Java

public boolean isSymmetric(TreeNode [1], TreeNode [2]) {
    if ([1] == null && [2] == null) return true;
    if ([1] == null || [2] == null) return false;
    if ([1].val != [2].val) return false;
    return isSymmetric([1].left, [2].right) && isSymmetric([1].right, [2].left);
}

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

Aroot1

Broot2

Cnode1

Dnode2

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing variable names between parameters and inside the method.

Using root1 and root2 inconsistently.

The helper method uses node1 and node2 as parameters and variables to compare nodes recursively.

Practice

1. What does it mean for two binary trees to be symmetric?

easy

A. They have identical left subtrees only.

B. They have the same number of nodes.

C. They are mirror images of each other around their center.

D. They have the same root value 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 C

4. Quick Check:

   Symmetry means mirror image = D [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 Java method signature correctly defines a function to check if two trees are symmetric?

easy

A. public boolean isSymmetric(TreeNode root1, TreeNode root2)

B. boolean isSymmetric(TreeNode root1, TreeNode root2) {}

C. public void isSymmetric(TreeNode root1, TreeNode root2)

D. public boolean isSymmetric(TreeNode root)

Solution

1. Step 1: Identify correct return type and parameters

   The method should return boolean and accept two TreeNode parameters for two trees.

2. Step 2: Check method signature syntax

   Public access, boolean return, and parameter types match in public boolean isSymmetric(TreeNode root1, TreeNode root2).

3. Final Answer:

   public boolean isSymmetric(TreeNode root1, TreeNode root2) -> Option A

4. Quick Check:

   Correct signature returns boolean with two TreeNode inputs [OK]

Hint: Boolean return with two TreeNode inputs is needed [OK]

Common Mistakes:

• Using void return type
• Missing public keyword
• Using only one TreeNode parameter

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

TreeNode root1 = new TreeNode(1);
root1.left = new TreeNode(2);
root1.right = new TreeNode(3);
TreeNode root2 = new TreeNode(1);
root2.left = new TreeNode(3);
root2.right = new TreeNode(2);

System.out.println(isSymmetric(root1, root2));

public static boolean isSymmetric(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null || t2 == null) return false;
    return (t1.val == t2.val) && isSymmetric(t1.left, t2.right) && isSymmetric(t1.right, t2.left);
}

medium

A. Runtime exception

B. false

C. Compilation error

D. true

Solution

1. Step 1: Analyze tree structure and values

   root1 has left=2, right=3; root2 has left=3, right=2, which are mirror images.

2. Step 2: Trace recursive calls

   Values match and recursive calls check mirrored children, all return true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Mirror structure and values = true [OK]

Hint: Check mirrored nodes and values recursively [OK]

Common Mistakes:

• Ignoring mirrored child comparison
• Confusing left and right children
• Assuming same structure means false

4. Identify the bug in this code snippet that checks if two trees are symmetric:

public boolean isSymmetric(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null || t2 == null) return false;
    return (t1.val == t2.val) && isSymmetric(t1.left, t2.left) && isSymmetric(t1.right, t2.right);
}

medium

A. It compares left with left and right with right instead of mirrored children.

B. It does not check for null nodes.

C. It returns true when both nodes are null.

D. It uses incorrect return type.

Solution

1. Step 1: Review recursive calls for symmetry

   The code compares t1.left with t2.left and t1.right with t2.right, which checks identical structure, not mirror.

2. Step 2: Identify correct mirrored comparison

   It should compare t1.left with t2.right and t1.right with t2.left for symmetry.

3. Final Answer:

   It compares left with left and right with right instead of mirrored children. -> Option A

4. Quick Check:

   Mirror check needs cross child comparison [OK]

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

Common Mistakes:

• Comparing same side children instead of mirrored
• Ignoring null checks
• Wrong return type usage

5. Given two binary trees, write a Java method to check if they are symmetric. Which approach correctly solves this problem? A) Traverse both trees in preorder and compare node values in order. B) Recursively check if left subtree of first tree is mirror of right subtree of second tree and vice versa. C) Compare the height of both trees and return true if equal. D) Check if both trees have the same number of leaf nodes.

hard

A. Traverse both trees in preorder and compare node values in order.

B. Recursively check if left subtree of first tree is mirror of right subtree of second tree and vice versa.

C. Compare the height of both trees and return true if equal.

D. Check if both trees have the same number of leaf nodes.

Solution

1. Step 1: Understand symmetry check requirements

   Symmetry means one tree is mirror image of the other, so left subtree of one matches right subtree of other.

2. Step 2: Evaluate approaches

   Recursively check if left subtree of first tree is mirror of right subtree of second tree and vice versa. correctly uses recursive mirror checks; others ignore structure or order.

3. Final Answer:

   Recursively check if left subtree of first tree is mirror of right subtree of second tree and vice versa. -> Option B

4. Quick Check:

   Mirror recursive check solves symmetry [OK]

Hint: Mirror recursive check is key to symmetry [OK]

Common Mistakes:

• Using preorder traversal comparison
• Comparing only height or leaf counts
• Ignoring mirrored subtree checks