Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following is not:
1 / \ 2 2 \ \ 3 3
Recursive: Pre-order comparison
class Solution { public: bool isSymmetric(TreeNode* left, TreeNode* right) { if(!left || !right) return left == right; return left->val == right->val && isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left); } bool isSymmetric(TreeNode* root) { return isSymmetric(root, root); } };