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)
        {
            TreeNode *t=new TreeNode(0);
            if(t1!=nullptr&&t2!=nullptr)
            {
                t->val=t1->val+t2->val;
                t->left=mergeTrees(t1->left, t2->left);
                t->right=mergeTrees(t1->right, t2->right);
            }
            else if(t1==nullptr){
                t->val=t2->val;
                t->left=mergeTrees(nullptr, t2->left);
                t->right=mergeTrees(nullptr, t2->right);
            }
            else {
                t->val=t1->val;
                t->left=mergeTrees(t1->left, nullptr);
                t->right=mergeTrees(t1->right, nullptr);
            }
            
            return t;
        }
        else {
            return nullptr;
        }
    }
};