题目地址：https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/
题目描述
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.)
Example 1:
Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
Note:
1、 1<=preorder.length<=100；
2、 Thevaluesofpreorderaredistinct.；
题目大意
给出了一个BST的先序遍历，求该BST。
解题方法
递归
先序遍历一定先遍历了根节点，所以出现的第一个数字一定是根。那么BST的左子树都比根节点小，而先序遍历要把左子树遍历结束才遍历右子树，所以向后找第一个大于根节点数字位置，该位置就是右子树的根节点。
做一个递归即可。
Python代码如下：
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def bstFromPreorder(self, preorder):
"""
:type preorder: List[int]
:rtype: TreeNode
"""
if not preorder: return None
root = TreeNode(preorder[0])
N = len(preorder)
i = 1
while i < N:
if preorder[i] > preorder[0]:
break
i += 1
root.left = self.bstFromPreorder(preorder[1:i])
root.right = self.bstFromPreorder(preorder[i:])
return root
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