Symmetric Tree
An easy Binary Trees problem included in Apna College, Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Binary Trees
- Sheets
- 3
- Core for
- 4 roles
- Platform
- LeetCode
The problem
Given the root of a binary tree, check whether it is a mirror of itself. A tree is symmetric if it looks the same when reflected around its center.
Example 1
- Input
- [1,2,2,3,4,4,3]
- Output
- true
- Why
- The left subtree is a mirror reflection of the right subtree.
Example 2
- Input
- [1,2,2,null,3,null,3]
- Output
- false
- Why
- The left subtree has 3 on the right side but the right subtree has 3 on the left side, so they are not mirrors.
Example 3
- Input
- [1]
- Output
- true
- Why
- A single node is always symmetric.
Constraints
- The number of nodes in the tree is in the range [1, 1000].
- -100 <= Node.val <= 100
How to think about it
Updated 2026-09-09Symmetry is not an operation on one tree; it is an equality comparison between two trees moving in opposite directions. The left child of the left branch must match the right child of the right branch, and the inner child of the left must match the inner child of the right.
Approaches, worst first
Iterative paired queue
time O(n) · space O(n)
Enqueue root's left and right children. Each iteration pops two nodes and verifies values. Then enqueue outer children together and inner children together. Avoids call stack recursion at the cost of queue overhead.
Dual recursive mirror checkWrite this one
time O(n) · space O(h)
Write a helper `isMirror(t1, t2)`. Both null returns true; exactly one null or value mismatch returns false. Recurse simultaneously on `(t1.left, t2.right)` and `(t1.right, t2.left)`. Minimal code with immediate early exit.
Where people lose marks · 3
- Comparing `t1.left` with `t2.left` instead of `t2.right`; standard equality checks identity rather than mirror reflection.
- Assuming symmetric inorder traversal guarantees a symmetric tree; a non-symmetric tree like [1,2,2,2,null,2] can share the same inorder values.
- Dereferencing `.left` or `.right` on a null pointer when one child is absent.
The theory behind it
Binary Trees — the ground this problem stands on. All Binary Trees problems
What Binary Trees is
A binary tree is a branching data structure that starts at a single top node called the root, like an upside-down family tree. Every node holds a piece of data and can branch out to at most two children below it, known as the left child and the right child. Because there is no ordering rule about which values go left or right, finding a specific item can require checking every single node in the entire tree.
When to reach for it
Reach for binary trees when problems present hierarchical data with left and right child pointers. Questions asking for tree height, maximum depth, path sums from root to leaf, diameter, lowest common ancestor, or checking whether two trees are mirror reflections of each other all signal binary tree traversals. Any problem asking to inspect or reconstruct a tree layer by layer or path by path belongs here.
How the pattern works
Think recursively by focusing on what a single node must do. If the current node is null, return the base answer immediately. Otherwise, ask the left child for its result, ask the right child for its result, and combine both answers with the current node value before returning up to the parent. For horizontal scans, use a queue to read nodes layer by layer, measuring the queue length at the start of each layer to group nodes by depth.
What each operation costs
| Operation | Time |
|---|---|
| traverse all nodes using recursion or queue | O(n) |
| search for an arbitrary value in an unordered tree | O(n) |
| call stack memory on balanced tree | O(log n) |
| call stack memory on skewed tree | O(n) |
What usually goes wrong with Binary Trees
- Dereferencing left or right child pointers without checking if the current node is null, throwing null pointer errors on empty trees or leaf nodes.
- Defining a leaf node incorrectly by stopping when either child is null instead of checking that both left and right children are simultaneously null.
- Computing tree diameter by taking left height plus right height inside a recursive helper without updating a global maximum across every visited node.
Which roles need this problem
Binary Trees is a core topic for these 4 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Full-Stack Developer, Android Developer, iOS Developer.
Track this in your role's order
Pick your target role and all 370 problems — including this one — resequence to what that interview actually asks. Free.
Start freeMore Binary Trees problems
- Count Good Nodes in Binary TreeMedium
- Balanced Binary TreeEasy
- Diameter of Binary TreeEasy
- Binary Tree Maximum Path SumHard
- Construct Binary Tree from Preorder and InorderMedium
- Construct Binary Tree from Inorder and PostorderMedium
- Serialize and Deserialize Binary TreeHard
- Lowest Common Ancestor of Binary TreeMedium
Problem set and role mapping as of .