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]
Note:
1 <= preorder.length <= 100
The values of preorder are distinct.
```

Solution:

```
func bstFromPreorder(preorder []int) *TreeNode {
lenP := len(preorder)
root := &TreeNode{Val: preorder[0]}
list := []*TreeNode{}
list = append(list, root)
for i:= 1; i < lenP; i++ {
node := &TreeNode{Val: preorder[i]}
pre := list[len(list) - 1]
if pre.Val > node.Val {
pre.Left = node
list = append(list, node)
continue
}
for list[len(list)-1].Val < node.Val {
if(len(list)> 1) {
pre = (list[len(list)-1])
list = list[0:len(list)-1]
}else {
pre = list[0]
list = list[0:0]
break
}
}
pre.Right = node;
list = append(list, node)
}
return root
}
```