# Increasing Order Search Tree Problem & Solution

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.

## C++ Solution

``````#pragma GCC optimize("Ofast")
#pragma GCC optimization("max-inline-insns-recursive-auto")
#pragma GCC target("avx,avx2,fma")

static const int _=[](){std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

/**
* 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* increasingBST(TreeNode* root) {
TreeNode* dummy = new TreeNode();
TreeNode* current = dummy;
dfs(root, current);

current = dummy->right;
delete dummy;

return current;
}

private:
void dfs(TreeNode* node, TreeNode*& current) {
if (node == nullptr) {
return;
}

dfs(node->left, current);

current->right = new TreeNode(node->val);
current = current->right;

dfs(node->right, current);
}
};
``````