Flatten Binary Tree to Linked List

题目：将二叉树（前序遍历的结果）转换成链表，不允许采用额外的空间

思想：将二叉树的右子树连接到左子树上面，然后将左子树连接到根的右边子树上，同时把根的左指针设置空

代码如下：

/**
 *  Definition for binary tree
 *  struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 *  };
 */
class Solution {
public:
     void flatten(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
           if(root == NULL)
           return ;
           flatten(root->left);
           flatten(root->right);
           if(root->left == NULL)
           return ;
           TreeNode *p = root;
           p = p->left;
           while(p->right)
           p = p->right;
           p->right = root->right;
           root->right = root->left;
           root->left = NULL;
    }
};