Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1
        / \
       2   5
      / \   \
     3   4   6

The flattened tree should look like:

   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

click to show hints.

Hints:

If you notice carefully in the flattened tree, each node’s right child points to the next node of a pre-order traversal.

题目解析：

这棵树的先序遍历是有序的。并且根据结构的特性，变形后，按照右子树链形成链表形式。

一般对树进行操作，也都用递归的形式。将做子树递归后形成的链表，插入到根结点和右子树之间。这样递归往复进行即可。

一定要有递归和模块的思想。就假设左边递归后已经形成，之后就按照链表处理就行了。这种方法比其他方法简单。

class Solution {
public:
    void flatten(TreeNode *root) {
        if(root == NULL)
            return ;
        if(root->left){
            flatten(root->left);
            TreeNode *cur = root->left;
            while(cur->right){
                cur = cur->right;
            }
            cur->right = root->right;
            root->right = root->left;
            root->left = NULL;
        }
        if(root->right)
            flatten(root->right);
    }
};