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);
    }
};