# Increasing Order Search Tree Problem

## Description

LeetCode Problem 897.

Given the root of 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 one right child.

Example 1:

``````1
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Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
``````

Example 2:

``````1
2
Input: root = [5,1,7]
Output: [1,null,5,null,7]
``````

Constraints:

• The number of nodes in the given tree will be in the range [1, 100].
• 0 <= Node.val <= 1000

## Sample C++ Code

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/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode *curr;
TreeNode* increasingBST(TreeNode* root) {
TreeNode* ans = new TreeNode(0);
curr = ans;
inorder(root);
return ans->right;
}

void inorder(TreeNode* node) {
if (node == NULL)
return;
inorder(node->left);
node->left = NULL;
curr->right = node;
curr = node;
inorder(node->right);
}
};
``````