LeetCode-in-All
101. Symmetric Tree
Easy
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
Follow up: Could you solve it both recursively and iteratively?
Solution
#include <stdio.h>
#include <stdbool.h>
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
// Helper function to check if two subtrees are mirror images
bool helper(struct TreeNode* leftNode, struct TreeNode* rightNode) {
if (leftNode == NULL || rightNode == NULL) {
return leftNode == NULL && rightNode == NULL;
}
if (leftNode->val != rightNode->val) {
return false;
}
return helper(leftNode->left, rightNode->right) && helper(leftNode->right, rightNode->left);
}
// Main function to check if the tree is symmetric
bool isSymmetric(struct TreeNode* root) {
if (root == NULL) {
return true;
}
return helper(root->left, root->right);
}