Increasing Order Search Tree

Description

Given a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9  

Note:

1. The number of nodes in the given tree will be between 1 and 100.

2. Each node will have a unique integer value from 0 to 1000.

Solution

class Solution {
public:
    TreeNode *increasingBST(TreeNode *root) {
        if (!root)
            return nullptr;
        return increasingBST(root, nullptr);
    }

    TreeNode *increasingBST(TreeNode *root, TreeNode *right) {
        auto head = root->left ? increasingBST(root->left, root) : root;
        root->left = nullptr;
        root->right = root->right ? increasingBST(root->right, right) : right;
        return head;
    }
};