Flatten Binary Tree to Linked List <leetcode>

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

算法：该题用的是递归，只要考虑当前节点，思路如下，设当前节点为root，把root的左节点的最右节点的右子树指向root的右子树，然后把root的右指针指向root的左子树，root的做指针置空，下一个节点处理root->right,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) {
        if(NULL==root)  return;
        if(NULL!=root->left)
        {
            getLeftRight(root)->right=root->right;
            root->right=root->left;
            root->left=NULL;
        }
        flatten(root->right);
        
    }
    
    TreeNode* getLeftRight(TreeNode* root)
    {
        root=root->left;
        while(NULL!=root->right)  root=root->right;
        return root;
    }
};