題目如下：

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.

解題思路：以用例的輸入[8,5,1,7,10,12]為例，很顯然8是根節點，8的左子樹有[5,1,7]，右子樹右[10,12]，左右子樹的分割點是後面第一個比根節點大的數。接下來再分別對[5,1,7]和[10,12]做同樣的操作，可以知道5是8的左子樹根節點，1和7分別在其左右；而10是8的右子樹根節點，12為右子樹節點。很顯然這是一個遞歸的過程，只要找到每個子樹的根節點將其左右子樹劃分即可。

代碼如下：

# 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 build(self,node,preorder):
        if len(preorder) == 0:
            return
        left 
 
= []
        for i in range(len(preorder)):
            if node.val < preorder[i]:
                break
            else:
                left.append(preorder[i])
        right = preorder[len(left):]
        if len(left) >= 1:
            node.left = TreeNode(left.pop(0))
            self.build(node.left,left)
        if len(right) >= 1:
            node.right = TreeNode(right.pop(0))
            self.build(node.right,right)
    def bstFromPreorder(self, preorder):
        """
        :type preorder: List[int]
        :rtype: TreeNode
        """
        root = TreeNode(preorder.pop(0))
        self.build(root,preorder)
        return root

【leetcode】1008. Construct Binary Search Tree from Preorder Traversal