LeetCode 101: Symmetric Tree — Python Solution

LeetCode 101: Symmetric Tree is an Easy binary tree general problem. This Python walkthrough develops a mirror recursion solution, ties every code decision to a concrete invariant, and includes the regression check used before publication.

This independent guide paraphrases the task rather than reproducing LeetCode’s prompt. Use the official page for the exact statement, examples, constraints, and submission runner.

Recognizing the pattern

Tree recursion works when the return value has one precise meaning for every subtree; the parent can then combine child results locally.

For this problem specifically, compare the outer and inner children of paired nodes: left with right and right with left. The invariant worth writing beside the code is: Each recursive pair must occupy mirrored positions around the root.

Step-by-step algorithm

1. Identify the input state consumed by isSymmetric(root) and initialize the data required by the mirror recursion pattern.
2. Compare the outer and inner children of paired nodes: left with right and right with left.
3. After each update, verify the page’s central invariant: Each recursive pair must occupy mirrored positions around the root.
4. Finish only after the boundary behavior is covered: Two missing mirrored nodes match; exactly one missing node fails.

Python solution

LeetCode supplies the list, tree, or graph node class referenced by this method. The downloadable test suite includes compatible local node definitions so the implementation can also run outside the judge.

class Solution:
    def isSymmetric(self, root):
        def mirror(left, right):
            if not left or not right:
                return left is right
            return left.val == right.val and mirror(left.left, right.right) and mirror(left.right, right.left)
        return mirror(root.left, root.right) if root else True

Reading the implementation

The main entry point is isSymmetric(root). The implementation keeps little named state because each operation can be resolved directly from the current input position.

The method expresses the transformation directly without a general traversal loop. Early returns stop as soon as the answer is forced, avoiding work that cannot change the result. In concrete terms, compare the outer and inner children of paired nodes: left with right and right with left.

Correctness argument

Base case. Empty or terminal subproblems return a result that already satisfies the claim for that smallest state.

Inductive step. Compare the outer and inner children of paired nodes: left with right and right with left. Assuming child or smaller states are correct, the current call combines only results allowed by the problem, so each recursive pair must occupy mirrored positions around the root.

Conclusion. Every recursive call reduces the remaining state. The base cases terminate, and induction carries the invariant back to the original input, establishing the returned answer.

Complexity and trade-offs

O(n) time and O(h) recursion space. The auxiliary-space figure excludes the returned output unless the output itself is the structure being built.

An explicit stack avoids recursion-depth limits, but it must carry the same state that recursive call frames provide automatically. That comparison is useful in an interview because it explains why the final implementation is preferable, not merely that it passes.

Regression check

The published implementation belongs to a 100-problem suite that is compiled and exercised behaviorally before deployment.

One reference assertion from that suite is shown below. It targets the normal path while the edge conditions in the next section cover the failure-prone boundaries.

assert Solution().isSymmetric(tree([1,2,2,3,4,4,3]))

Common mistakes and edge cases

• Problem-specific boundary: Two missing mirrored nodes match; exactly one missing node fails.
• Pattern-level pitfall: Write the empty-subtree result first and keep returned subtree information separate from any global answer updated at a node.
• Invariant check: after every update, confirm that each recursive pair must occupy mirrored positions around the root.

Interview review checklist

• Explain why mirror recursion matches the structure of this input.
• State the invariant in one sentence before tracing code: Each recursive pair must occupy mirrored positions around the root.
• Derive O(n) time and O(h) recursion space from how many times each element or state is visited.
• Test the boundary explicitly: Two missing mirrored nodes match; exactly one missing node fails.

Browse the searchable 100 LeetCode Python Solutions hub. Previous: 226. Invert Binary Tree · Next: 105. Construct Binary Tree from Preorder and Inorder Traversal