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Abhishek Chaudhary

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# Construct Binary Search Tree from Preorder Traversal

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.

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 getRoot(self, p, q):
return min((i for i in range(p, q)), key = lambda x: self.valIndex[self.inorder[x]])

def buildTreeRec(self, p, q):
if p < q:
root = TreeNode()
currRoot = self.getRoot(p, q)
root.val = self.inorder[currRoot]
if currRoot > p:
root.left = TreeNode()
root.left = self.buildTreeRec(p, currRoot)
if currRoot < q - 1:
root.right = TreeNode()
root.right = self.buildTreeRec(currRoot + 1, q)
return root

def bstFromPreorder(self, preorder: List[int]) -> Optional[TreeNode]:
n = len(preorder)
self.inorder = sorted(preorder)
self.valIndex = {}
for i, item in enumerate(preorder):
self.valIndex[item] = i
return self.buildTreeRec(0, n)
``````