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]
Solution
利用二叉搜索树的特点，它的左子树小于根节点，右子树大于根节点。二叉树的先序遍历，都有一个元素是根节点，然后比根节点小的都是它的左子树，剩下的就是它的左子树，进行递归调用，直到preorder为空。
# 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:
if not preorder:
return
val = preorder[0]
index =1
while index < len(preorder) and preorder[index]<val:
index = index+1
root = TreeNode(val)
root.left = self.bstFromPreorder(preorder[1:index])
root.right = self.bstFromPreorder(preorder[index:])
return root
