leetcode-1008-construct-binary-search-tree-from-preorder-traversal.swift

/**

Copyright (c) 2020 David Seek, LLC

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https://leetcode.com/problems/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.

*/

/**
Big O Annotation
Time complexity O(n) where n is the amount of elements in preorder.
Space complexity O(1) we're not using any extra space.
*/
func bstFromPreorder(_ preorder: [Int]) -> TreeNode? {
        
    /**
    If there's nothing in the array, 
    there's not gonna be a BST
    */
    guard !preorder.isEmpty else {
        return nil
    }
    
    // Get the root node
    let root = TreeNode(preorder[0])
    
    // Fill it with the remaining values...
    for i in 1..<preorder.count {
        root.insert(preorder[i])
    }
    
    return root
}

extension TreeNode {

    // Simple insert function
    func insert(_ value: Int) {
        
        // If the value is smaller than root...
        if value < val {
            
            // Check if we already have a left node...
            if let left = left {

                // If so, start recursion
                left.insert(value)
            } else {
                
                // If not, make it the left node
                left = TreeNode(value)
            }
            
        } else {
            
            // Same on the right
            if let right = right {
                right.insert(value)
            } else {
                right = TreeNode(value)
            }
        }
    }
}