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Symmetric Tree – Solution & Complexity

Interview solutions
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Symmetric Tree

Solution Walkthrough

1. Recognize the Pattern

The key pattern is: Parallel DFS that mirrors the left/right subtrees against each other. Identify the state and invariant before coding.

2. Build the Algorithm

Advance one state transition at a time. Mark or update state before exploring dependent work.

3. Check Edge Cases

Test empty or minimal input, skewed shapes, duplicates where allowed, and impossible outcomes.

4. Solution and Complexity

Time: O(n). Space: O(h) for the recursion stack (O(n) worst case on a skewed tree).

All 5 languages below run and submit against the remote judge for this problem — Java/Go/Rust use the same real TreeNode structural-type support the judge added for the rest of the linked-list/tree track.

def is_symmetric(root: TreeNode) -> bool:
    def mirror(t1, t2):
        if t1 is None and t2 is None:
            return True
        if t1 is None or t2 is None:
            return False
        return (
            t1.val == t2.val
            and mirror(t1.left, t2.right)
            and mirror(t1.right, t2.left)
        )

    return mirror(root, root)

FAQ

See also

Practice Symmetric Tree online (run code in the browser)Diameter of Binary Tree solution (easy)Invert Binary Tree solution (easy)Maximum Depth of Binary Tree solution (easy)Same Tree solution (easy)All coding interview problems
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