[LeetCode] 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.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct Node
{
    TreeNode *head;
    TreeNode *tail;
    Node(){}
    Node(TreeNode *h, TreeNode *t):head(h), tail(t){}
};

class Solution {
public:
    Node convert(TreeNode *root)
    {
        if (root == NULL)
            return Node(NULL, NULL);
            
        Node leftNode = convert(root->left);
        Node rightNode = convert(root->right);
        
        root->right = NULL;
        root->left = NULL;
        
        if (leftNode.head)
        {
            root->right = leftNode.head;
            leftNode.tail->right = rightNode.head;
            
        }
        else
        {
            root->right = rightNode.head;
        }
        
        TreeNode *tail;
        if (rightNode.tail != NULL)
            tail = rightNode.tail;
        else if (leftNode.tail != NULL)
            tail = leftNode.tail;
        else
            tail = root;
            
        return Node(root, tail);
    }
    
    void flatten(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        convert(root);
    }
};