[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.

 

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 struct Node
11 {
12     TreeNode *head;
13     TreeNode *tail;
14     Node(){}
15     Node(TreeNode *h, TreeNode *t):head(h), tail(t){}
16 };
17 
18 class Solution {
19 public:
20     Node convert(TreeNode *root)
21     {
22         if (root == NULL)
23             return Node(NULL, NULL);
24             
25         Node leftNode = convert(root->left);
26         Node rightNode = convert(root->right);
27         
28         root->right = NULL;
29         root->left = NULL;
30         
31         if (leftNode.head)
32         {
33             root->right = leftNode.head;
34             leftNode.tail->right = rightNode.head;
35             
36         }
37         else
38         {
39             root->right = rightNode.head;
40         }
41         
42         TreeNode *tail;
43         if (rightNode.tail != NULL)
44             tail = rightNode.tail;
45         else if (leftNode.tail != NULL)
46             tail = leftNode.tail;
47         else
48             tail = root;
49             
50         return Node(root, tail);
51     }
52     
53     void flatten(TreeNode *root) {
54         // Start typing your C/C++ solution below
55         // DO NOT write int main() function
56         convert(root);
57     }
58 };