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.

Solution: Recursion. Return the last element of the flattened sub-tree.

/**
 * 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) {
        TreeNode* end = NULL;
        flattenRe(root, end);
    }
    
    void flattenRe(TreeNode* root, TreeNode* &end)
    {
        if(!root) return;
        TreeNode* lend = NULL;
        TreeNode* rend = NULL;
        flattenRe(root->left, lend);
        flattenRe(root->right, rend);
        if(root->left) {
            lend->right = root->right;
            root->right = root->left;
            root->left = NULL;
        }
        end = rend ? rend : (lend ? lend : root);
    }
};