How do you solve Symmetric Tree on LeetCode?

Symmetric Tree (LeetCode #101) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: Symmetry can be checked directly by comparing nodes

What is the brute force solution for Symmetric Tree?

The brute force approach for Symmetric Tree is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach checks for symmetry by comparing each node with its corresponding node in the opposite subtree. This is done by traversing the tree multiple times, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Symmetric Tree?

Symmetric Tree uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Recursion, Tree Traversal.

What are the interview tips for Symmetric Tree?

Explain your thought process clearly Consider edge cases like single-node trees Practice both recursive and iterative approaches

What are common mistakes when solving Symmetric Tree?

Not handling null nodes properly Confusing left and right subtree comparisons

#101

Symmetric Tree

Easy

TreeDepth-First SearchBreadth-First SearchBinary TreeRecursionTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(h)

💡

Intuition

Time O(n)Space O(h)

The optimal approach uses a recursive helper function to directly compare the left and right subtrees. This is efficient because it avoids extra space for storing node values and checks symmetry in a single traversal.

⚙️

Algorithm

4 steps

1Step 1: Create a helper function that takes two nodes as parameters.
2Step 2: If both nodes are null, return true (base case).
3Step 3: If one node is null and the other is not, return false.
4Step 4: Check if the values of both nodes are equal, and recursively check the left and right children in a mirrored fashion.

solution.py17 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def isSymmetric(self, root: TreeNode) -> bool:
10        def isMirror(left, right):
11            if not left and not right:
12                return True
13            if not left or not right:
14                return False
15            return (left.val == right.val) and isMirror(left.left, right.right) and isMirror(left.right, right.left)
16
17        return isMirror(root, root)

ℹ

Complexity note: The time complexity is O(n) because we visit each node once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1Symmetry can be checked directly by comparing nodes
2Recursion can simplify tree traversal problems

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.