题解 | #合并二叉树#C++暴力遍历递归解法

/**
 * 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* answer(TreeNode*t1,TreeNode*t2)
    {
        if(t1==NULL  &&  t2==NULL)
        {
            return NULL;
        }
        else 
        {
            TreeNode*root;
            if(t1==NULL)
            {
                root=t2;
                return root;
            }
            else 
            {
                if(t2==NULL)
                {
                    root=t1;
                    return t1;
                }
                else 
                {
                    t1->val+=t2->val;
                    root=t1;
                    root->left=answer(t1->left,t2->left);
                    root->right=answer(t1->right,t2->right);
                    return root;
                }
            }
        }
    }
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) 
    {
        return answer(t1,t2);
    }
};