What is the brute force approach and complexity for Symmetric Tree?

Make an inverted copy of the whole tree, check it equals the original, then free the copy with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for Symmetric Tree?

Walk the left and right subtrees together, with the children crossed with optimal time complexity O(N) and space complexity O(H).

How do you identify the pattern for Symmetric Tree?

Look for key signals in the prompt: "mirror of itself" indicates Compare the left side with the right side, crossed: the left of one with the right of the other.; "symmetric around its center" indicates Both the outer pair and the inner pair must mirror. Each side on its own does not decide the answer.; "one root, two children" indicates The check starts at the root's two children, which form the first mirror pair..

Symmetric Tree

Easy·LeetCode #101 Pro OnlyMedium

LeetCode

Problem Statement

You are given the root of a binary tree. Decide whether the tree is a mirror of itself: if you flip it left to right around its center, it looks exactly the same.

Return true if the tree is symmetric, otherwise false.

Input

The root of a binary tree (it has at least one node).

Output

true if the tree is symmetric, false otherwise.

Example 1

Input: root = [1, 2, 2, 3, 4, 4, 3]

Output: true

The left subtree (2 with children 3 and 4) and the right subtree (2 with children 4 and 3) are mirror images of each other.

Example 2

Input: root = [1, 2, 2, null, 3, null, 3]

Output: false

Both 3s are right children. A mirror image would put one of them on the left, so the two sides have different shapes.

Constraints

• The number of nodes in the tree is in the range [1, 1000].
• −100 ≤ Node.val ≤ 100

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“mirror of itself”

→ Compare the left side with the right side, crossed: the left of one with the right of the other.

“symmetric around its center”

→ Both the outer pair and the inner pair must mirror. Each side on its own does not decide the answer.

“one root, two children”

→ The check starts at the root's two children, which form the first mirror pair.

Approach Complexity Ladder

Brute Force — build a mirror copy and compareO(N)O(N)

Make an inverted copy of the whole tree, check it equals the original, then free the copy

Optimal — compare the two sides as mirror pairsO(N)O(H)

Walk the left and right subtrees together, with the children crossed

Brute Force Intuition

Build a second tree that is the mirror image of the first: at every node, swap the left and right children. A tree is symmetric exactly when it equals its own mirror image, so compare the two with the Same Tree check. The copy costs one new node for every node of the tree.

Recognition rule for this sub-pattern:

Visit nodes in pre/in/post order or walk two trees together node by node.

Remember this · Tree DFS

Trigger

Paths, heights, diameters, validating structure, ancestors, anything "for every subtree".

Think

If my left and right children already gave me their answers, how do I build mine?

Key question

What does my recursive function return, and what do I pass down?

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