617. 合并二叉树(c++)

暴力解，当t1为空返回t2，当t2为空返回t1。当t1，t2都有值，new新节点为两个节点的和，新节点左子树为原始节点左子树合并，新节点右子树为原始节点右子树合并。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (t1 == NULL)
        {
            return t2;
        }
        if (t2 == NULL)
        {
            return t1;
        }
        TreeNode* newnode = new TreeNode(t1->val + t2->val);
        newnode->left = mergeTrees(t1->left, t2->left);
        newnode->right = mergeTrees(t1->right, t2->right);
        return newnode;

    }
};