Flatten Binary Tree to Linked List

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

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

开始想的太简单了，很困，也没认真思考，直接看的别人怎么做的，这道题还是很考验思维的，解法和以前那些深度优先算法不一样，思考清楚了之后，题目也就好解了，很久没用递归解题了，导致都忘了可以用递归来解了，很多的关于树的题都可以考虑用递归来解·····································

代码如下：

/**
* 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:
void flatten(TreeNode* root) {
if(root==NULL)
return ;
if(root->left!=NULL)
{
TreeNode* leftpart=root->left;
TreeNode* rightpart=root->right;
root->left=NULL;
root->right=leftpart;
TreeNode* temp=root->right;
while(temp->right!=NULL)
{
temp=temp->right;
}
temp->right=rightpart;
}
flatten(root->right);
}
};