# 1008. Construct Binary Search Tree from Preorder Traversal 前序遍历构造二叉搜索树

@TOC

## # 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. The values of preorder are distinct.

## # 解题方法

### # 递归

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
``````

## # 日期

2019 年 3 月 10 日 —— 周赛进了第一页！