Day 20 of 30 days Leetcode challenge - Construct Binary Search Tree from Preorder Traversal

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)

Example 1:

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


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


Use recursive.
All values of left sub-tree will lower than value of root node, higher for right sub tree.
The trick is using Int.min, Int.max as first range.

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