# Construct Binary Search Tree from Preorder Traversal

## Description

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`.)_

**Example 1:**

```
Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
```

![](https://1237518168-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LH66aUbuEuCogtsbHMt%2F-Lo4VEmRx0hQkyLY63oe%2F-Lo4VUDKMNwSUSrtLZmc%2Fimage.png?alt=media\&token=0e8aeae9-1945-4be7-881f-78c2fd9e6539)

**Note:**&#x20;

1. `1 <= preorder.length <= 100`
2. The values of `preorder` are distinct.

## Solution

```cpp
class Solution {
public:
    TreeNode *bstFromPreorder(vector<int> &preorder) {
        if (preorder.empty())
            return nullptr;
        TreeNode *node, *curr, *root = new TreeNode(preorder[0]);
        stack<TreeNode *> st;
        st.push(root);
        for (int i = 1; i < preorder.size(); ++i) {
            int val = preorder[i];
            if (val < st.top()->val) {
                node = new TreeNode(val);
                st.top()->left = node;
                st.push(node);
            } else {
                do {
                    curr = st.top();
                    st.pop();
                } while (!st.empty() && st.top()->val < val);
                curr->right = new TreeNode(val);
                st.push(curr->right);
            }
        }
        return root;
    }
};
```
