题解 | #合并二叉树#

按照t1先序遍历

若t1和t2都不为空，那么t1和t2的val相加，递归help(t1->left,t2->left)和help(t1->right,t2->right)

若t1为空，t2不为空，令t1=t2

若t1不为空，t2为空，不用操作

若t1、t2都为空，直接返回即可

 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param t1 TreeNode类 
     * @param t2 TreeNode类 
     * @return TreeNode类
     */
    void help(TreeNode* &t1,TreeNode* t2){
        if(t1==NULL&&t2==NULL) return ;
        else if(t1==NULL&&t2!=NULL) t1=t2;
        else if(t1!=NULL&&t2!=NULL) t1->val+=t2->val,help(t1->left,t2->left),help(t1->right,t2->right);
        
    }
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        // write code here
        help(t1,t2);
        return t1;
    }
};