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==NULL && t2 ==NULL){
            return NULL;
        }
        if(t1==NULL) return t2;
        if(t2==NULL) return t1;

        t1->val = t1->val +t2->val;
        t1->left = mergeTrees(t1->left, t2->left);
        t1->right = mergeTrees(t1->right, t2->right); 
        return t1;

    }
};

还是很精巧，需要思考，先都往左走，走不动了就看谁还在，在的t1的左，

右边同理。做t1的右

如果都有就求sum