// https://leetcode.com/problems/symmetric-tree/submissions/1292981419/

/**
 * Definition for a binary tree node.
*/
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

/**
 * Method:
 *  - Reverse the right-hand side
 *  - Check whether the left and right are now equal via a pre-order traversal.
 *  - Time: O(n)
 *  - Space: O(n).
 *
 * On my first time, I didn't see that the problem could be solved directly by comparing the trees recursively;
 * doing this would still be O(n) time and space, but avoid having to invert the right tree for no other reason.
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        reverseBinaryTree(root->right);
        return treesAreEqual(root->left, root->right);
    }

    void reverseBinaryTree(TreeNode* root) {
        if (root == nullptr) return;
        TreeNode* temp = root->left;
        root->left = root->right;
        root->right = temp;
        reverseBinaryTree(root->left);
        reverseBinaryTree(root->right);
    }

    bool treesAreEqual(TreeNode* left, TreeNode* right) {
        if (left == nullptr) return right == nullptr;
        if (right == nullptr) return left == nullptr;
        return (left->val != right->val) && treesAreEqual(left->left, right->left) && treesAreEqual(left->right, right->right);
    }
};