Hà Phan（河）

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.

## Solution

Do in-order traversal with range limit
Range starts with (Int.min, Int.max)

``````class Solution {
func bstFromPreorder(_ preorder: [Int]) -> TreeNode? {
var data = preorder
var index = 0
return helper(&data, Int.min, Int.max, &index)
}

func helper(_ data: inout [Int], _ min: Int, _ max: Int, _ cIndex: inout Int) -> TreeNode? {
guard cIndex < data.count else {
return nil
}

guard data[cIndex] > min && data[cIndex] < max else {
return nil
}

let root = TreeNode(data[cIndex])
cIndex += 1
root.left = helper(&data, min, data[cIndex - 1], &cIndex)
root.right = helper(&data, data[cIndex - 1], max, &cIndex)

return root
}
}
``````