What if you could instantly tell if two complex trees are perfect mirror images without getting lost in details?
Why Check if Two Trees are Symmetric in DSA Typescript?
Imagine you have two family trees drawn on paper, and you want to see if one is a mirror image of the other. Doing this by looking at each branch and comparing them one by one can be confusing and tiring.
Manually checking each branch and leaf for symmetry is slow and easy to mess up. You might miss a small difference or get lost in the details, especially if the trees are big.
Using a simple method to compare the trees step-by-step automatically checks if they are mirror images. This saves time and avoids mistakes by following clear rules for symmetry.
function areSymmetricManual(tree1, tree2) {
// Manually compare nodes one by one
// Lots of repeated checks and confusion
}function areSymmetric(tree1, tree2) {
if (!tree1 && !tree2) return true;
if (!tree1 || !tree2) return false;
return tree1.val === tree2.val &&
areSymmetric(tree1.left, tree2.right) &&
areSymmetric(tree1.right, tree2.left);
}This lets you quickly and correctly find out if two trees are mirror images, no matter how big or complex they are.
Checking if two family trees or organizational charts are symmetrical helps spot patterns or errors in data entry.
Manual comparison is slow and error-prone.
Recursive checking automates symmetry verification.
Enables fast, reliable mirror-image detection in trees.
Practice
Solution
Step 1: Understand symmetry in trees
Two trees are symmetric if one is the mirror reflection of the other around the root node.Step 2: Differentiate from other properties
Having the same number of nodes or values in order does not guarantee symmetry; symmetry requires mirrored structure and values.Final Answer:
They are mirror images of each other around their root. -> Option AQuick Check:
Symmetry means mirror image = A [OK]
- Confusing symmetry with identical trees
- Thinking same node count means symmetry
- Assuming balanced trees are always symmetric
Solution
Step 1: Identify correct parameters and return type
The function should take two tree roots and return a boolean indicating symmetry.Step 2: Check function signature correctness
function isSymmetric(root1: TreeNode, root2: TreeNode): boolean matches this: two TreeNode parameters and boolean return type.Final Answer:
function isSymmetric(root1: TreeNode, root2: TreeNode): boolean -> Option BQuick Check:
Two roots + boolean return = D [OK]
- Using only one root parameter
- Returning number instead of boolean
- Using void return type
class TreeNode {
val: number;
left: TreeNode | null;
right: TreeNode | null;
constructor(val: number, left: TreeNode | null = null, right: TreeNode | null = null) {
this.val = val;
this.left = left;
this.right = right;
}
}
function isMirror(t1: TreeNode | null, t2: TreeNode | null): boolean {
if (!t1 && !t2) return true;
if (!t1 || !t2) return false;
return (t1.val === t2.val) && isMirror(t1.left, t2.right) && isMirror(t1.right, t2.left);
}
const tree1 = new TreeNode(1, new TreeNode(2), new TreeNode(3));
const tree2 = new TreeNode(1, new TreeNode(3), new TreeNode(2));
console.log(isMirror(tree1, tree2));Solution
Step 1: Understand the isMirror function logic
It checks if two trees are mirror images by comparing root values and recursively checking left of one with right of other and vice versa.Step 2: Trace the given trees
tree1 has root 1, left 2, right 3; tree2 has root 1, left 3, right 2. They are mirrors, so isMirror returns true.Final Answer:
true -> Option CQuick Check:
Mirror check returns true for these trees [OK]
- Confusing left and right child comparison
- Assuming same structure means mirror
- Ignoring null checks
function isSymmetric(t1: TreeNode | null, t2: TreeNode | null): boolean {
if (!t1 && !t2) return true;
if (!t1 || !t2) return false;
if (t1.val !== t2.val) return false;
return isSymmetric(t1.left, t2.left) && isSymmetric(t1.right, t2.right);
}Solution
Step 1: Analyze recursive calls
The function compares t1.left with t2.left and t1.right with t2.right, which checks identical structure, not mirror symmetry.Step 2: Identify correct mirror comparison
For symmetry, t1.left should be compared with t2.right and t1.right with t2.left.Final Answer:
It compares left with left and right with right instead of mirror sides. -> Option DQuick Check:
Mirror needs cross child comparison = A [OK]
- Comparing same side children instead of opposite
- Forgetting to check node values before recursion
- Missing null base cases
Solution
Step 1: Understand symmetry check requirements
Symmetry means the left subtree of one tree is the mirror of the right subtree of the other.Step 2: Choose correct approach
Recursively comparing left subtree of one with right subtree of the other ensures mirror symmetry is checked properly.Final Answer:
Use a helper function to recursively compare left subtree of one tree with right subtree of the other and vice versa. -> Option AQuick Check:
Mirror recursion approach = B [OK]
- Comparing preorder traversals which miss mirror structure
- Checking only height or node count
- Comparing only root values
Ready to prove it?
Pass the test and this pattern moves unknown
