[Solved] Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root. It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.

Question

Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.

It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.

A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.

A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.

Example 1:

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

Example 2:

Input: preorder = [1,3]
Output: [1,null,3]

Constraints:

• 1 <= preorder.length <= 100
• 1 <= preorder[i] <= 1000
• All the values of preorder are unique.

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def bstFromPreorder(self, preorder: List[int]) -> TreeNode:
        root = TreeNode(preorder[0])
        stack = [root]
        
        for i in preorder[1:]:
            if i<stack[-1].val:
                stack[-1].left = TreeNode(i)
                stack.append(stack[-1].left)
            else:
                while stack and stack[-1].val<i:
                    last = stack.pop()
                last.right = TreeNode(i)
                stack.append(last.right)
        
        return root