Construct Binary Search Tree from Preorder Traversal

Description

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(javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} preorder
 * @return {TreeNode}
 */
var bstFromPreorder = function(preorder) {
    let index = 0
    const aux = (bound = Infinity) => {
        if(index >= preorder.length || preorder[index] > bound) {
            return null
        }
        const node = new TreeNode(preorder[index])
        index++
        node.left = aux(node.val)
        node.right = aux(bound)
        return node
    }
    return aux()
};