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.

See the increasing order search tree problem on LeetCode.

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

    current = dummy->right;
    delete dummy;

    return current;

  void dfs(TreeNode* node, TreeNode*& current) {
    if (node == nullptr) {

    dfs(node->left, current);

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

    dfs(node->right, current);

Start Here

Many paths, there are. Follow yours, you must.