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							Symmetric Tree
								Certification: Basic Level
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								Submissions: 0
								Points: 5
							
							Write a C program to check if a binary tree is symmetric around its center, meaning it is a mirror of itself. A binary tree is symmetric if the left subtree of the root is a mirror reflection of the right subtree of the root.
Example 1
- Input:
 - Output: true
 - Explanation: 
- The left subtree of the root (1) consists of nodes 2, 3, and 4. 
 - The right subtree of the root (1) consists of nodes 2, 4, and 3. 
 - The left subtree is a mirror reflection of the right subtree. 
 - The values match in mirror positions (2-2, 3-3, 4-4). 
 - Therefore, the tree is symmetric.
 
 - The left subtree of the root (1) consists of nodes 2, 3, and 4. 
 
Example 2
- Input:
 - Output: false
 - Explanation: 
- The left subtree of the root (1) consists of nodes 2 and 3.
 - The right subtree of the root (1) consists of nodes 2 and 3. 
 - In the left subtree, node 3 is the right child of node 2. 
 - In the right subtree, node 3 is also the right child of node 2. 
 - In a mirror reflection, node 3 in the right subtree should be the left child of node 2. 
 - Since this is not the case, the tree is not symmetric.
 
 - The left subtree of the root (1) consists of nodes 2 and 3.
 
Constraints
- The number of nodes in the tree is in the range [1, 1000].
 - The values of nodes are integers in the range [-100, 100].
 - Time Complexity: O(n), where n is the number of nodes in the tree.
 - Space Complexity: O(h), where h is the height of the tree (due to recursion stack).
 
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Solution Hints
- Use a recursive approach to check if two subtrees are mirror images of each other.
 - Start by comparing the left and right subtrees of the root.
 - Two subtrees are mirror images if: a. Their root values are the same b. The left subtree of one is a mirror of the right subtree of the other c. The right subtree of one is a mirror of the left subtree of the other
 - Handle the base cases properly (e.g., when both nodes are NULL).
 - You can also use an iterative approach with a queue or stack if preferred.
 - Be careful about the comparison conditions to ensure proper mirroring.
 - Consider edge cases like empty trees or trees with only one node.