HàPhan 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`)

Example 1:

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

Note:

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

Solution

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.
https://www.geeksforgeeks.org/construct-bst-from-given-preorder-traversa/

``````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
}
}
``````