text: /*
  Construct Binary Search Tree from Preorder Traversal
  Return the root node of a binary search tree that matches the given preorder traversal.
  
  (Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val.  Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)
  
  It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.
  
  Example 1:
  
  Input: [8,5,1,7,10,12]
  Output: [8,5,10,1,7,null,12]
  
  Constraints:
  
  1 <= preorder.length <= 100
  1 <= preorder[i] <= 10^8
  The values of preorder are distinct.
  */
  
  /**
   * Definition for a binary tree node.
   * struct TreeNode {
   *     int val;
   *     TreeNode *left;
   *     TreeNode *right;
   *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
   * };
   */
  class Solution {
  public:
      TreeNode* bstFromPreorder(vector<int>& preorder) {
          j=preorder.size();
          return h(preorder, INT_MAX);
      }
  private:
      int i=0,j=0;
      TreeNode* h(vector<int>& preorder, int max){
          if(i==j||preorder[i]>max) return NULL;
          TreeNode *node = new TreeNode(preorder[i++]);
          // build the left tree taking the elements smaller than node->val
          node->left = h(preorder, node->val);
          // build the right tree taking the elements smaller than max
          node->right = h(preorder, max);
          return node;
      }
  };
  static const auto io_sync_off = []() {std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();