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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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In this post, you will find the solution for the **Symmetric Tree** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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- Mathematics/Basic Logical Based Questions
- Arrays
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- Dynamic Programming
- Stack & Queue
- Trees & Graphs
- Greedy Algorithms
- Breadth-First Search
- Depth-First Search
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- BST (Binary Search Tree)
- Database
- Linked List
- Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

** Link for the Problem** – Symmetric Tree– LeetCode Problem

Symmetric Tree– LeetCode Problem

**Problem:**

Given the `root`

of a binary tree, *check whether it is a mirror of itself* (i.e., symmetric around its center).

**Example 1:**

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

**Example 2:**

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

**Constraints:**

- The number of nodes in the tree is in the range
`[1, 1000]`

. `-100 <= Node.val <= 100`

Symmetric Tree– LeetCode Solutions

Symmetric Tree Solution in C++:

class Solution { public: bool isSymmetric(TreeNode* root) { return isSymmetric(root, root); } private: bool isSymmetric(TreeNode* p, TreeNode* q) { if (!p || !q) return p == q; return p->val == q->val && isSymmetric(p->left, q->right) && isSymmetric(p->right, q->left); } };

Symmetric Tree Solution in Java:

class Solution { public boolean isSymmetric(TreeNode root) { return isSymmetric(root, root); } private boolean isSymmetric(TreeNode p, TreeNode q) { if (p == null || q == null) return p == q; return p.val == q.val && isSymmetric(p.left, q.right) && isSymmetric(p.right, q.left); } }

Symmetric Tree Solution in Python:

class Solution: def isSymmetric(self, root: Optional[TreeNode]) -> bool: def isSymmetric(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool: if not p or not q: return p == q return p.val == q.val and \ isSymmetric(p.left, q.right) and \ isSymmetric(p.right, q.left) return isSymmetric(root, root)