code.cpp

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 *
 * Code for the book "DAILY 43: Algorithmic Problems for Coding Interviews: Easy level, C++ edition"
 *
 * Licensed under the GNU General Public License, Version 3.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *     http://www.gnu.org/licenses/gpl-3.0.html
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 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
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 *
 * For details about the book, please visit: https://www.amazon.com/dp/B0CZJNBLQS
 * *************************************************************************/

#include <iostream>

// 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) {}
};

class Solution {
public:
    bool isSameBinaryTree(TreeNode* p, TreeNode* q) {
        if (p == nullptr && q == nullptr) {
            return true;
        }
        if (p == nullptr || q == nullptr) {
            return false;
        }
        if (p->val == q->val) {
            return isSameBinaryTree(p->left, q->left) && 
                   isSameBinaryTree(p->right, q->right);
        }
        return false;
    }
};

int main() {
    // Example usage
    Solution sol;

    TreeNode* tree1 = new TreeNode(1);
    tree1->left = new TreeNode(2);
    tree1->right = new TreeNode(3);

    TreeNode* tree2 = new TreeNode(1);
    tree2->left = new TreeNode(2);
    tree2->right = new TreeNode(3);

    if (sol.isSameBinaryTree(tree1, tree2)) {
        std::cout << "The trees are the same." << std::endl;
    } else {
        std::cout << "The trees are not the same." << std::endl;
    }

    // Clean up
    delete tree1->left;
    delete tree1->right;
    delete tree1;
    delete tree2->left;
    delete tree2->right;
    delete tree2;

    return 0;
}