题解 | #合并二叉树#

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 *  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param t1 TreeNode类
     * @param t2 TreeNode类
     * @return TreeNode类
     */
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        // write code here
        if (t1 == nullptr && t2 == nullptr) {
            return nullptr;
        } else if (t1 == nullptr) {
            return t2;
        } else if (t2 == nullptr) {
            return t1;
        } else {
            t1->val += t2->val;
            if (t1->left == nullptr && t2->left != nullptr) {
                t1->left = t2->left;
            } else if (t1->left != nullptr && t2->left != nullptr) {
                mergeTrees(t1->left, t2->left);
            }
            if (t1->right == nullptr && t2->right != nullptr) {
                t1->right = t2->right;
            } else if (t1->right != nullptr && t2->right != nullptr) {
                mergeTrees(t1->right, t2->right);
            }
            return t1;
        }
    }
};