leetcode 617. Merge Two Binary Trees

这个题目本质是对两棵树做遍历，同时生产一棵新的树，树的节点值就是原来两棵树节点值的和

刚开始我的思路如下：

class Solution {public:    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {        if(t1==NULL && t2==NULL)        {            return NULL;        }        else        {            TreeNode *node=new TreeNode((t1?t1->val:0) + (t2?t2->val:0) );            node->left=mergeTrees((t1->left,t2->left);            node->right=mergeTrees(t1->right,t2->right);            return node;        }    }};

但是报错了，错误为：member access within null pointer of type 'struct TreeNode'  意思是访问了空指针中的成员变量，例如t1==NULL，但是我用到了t1->left和t2->right，这个很明显是错误的。因此改完之后如下：

class Solution {public:    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {        if(t1==NULL && t2==NULL)  //递归边界        {            return NULL;        }        else        {            TreeNode *node=new TreeNode((t1?t1->val:0) + (t2?t2->val:0) ); //这种用法以前很少见到，值得铭记            node->left=mergeTrees((t1?t1->left:NULL),(t2?t2->left:NULL));  //用三元表达式来选择传入哪种参数            node->right=mergeTrees(t1?t1->right:NULL,(t2?t2->right:NULL));            return node;        }    }};