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
8
/ \
5 10
/ \ \
1 7 12Input
preorder = [8,5,1,7,10,12]Output
[8,5,10,1,7,null,12]The constructed BST has root 8, left subtree rooted at 5 with children 1 and 7, and right subtree rooted at 10 with right child 12.
Example 2
1
\
3Input
preorder = [1,3]Output
[1,null,3]The preorder traversal [1, 3] constructs a BST with root 1 and right child 3.
Constraints
- 1 <= preorder.length <= 100
- 1 <= preorder[i] <= 1000
- All the values of
preorderare unique.