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

题目如下：

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.

解题思路：以用例的输入[8,5,1,7,10,12]为例，很显然8是根节点，8的左子树有[5,1,7]，右子树右[10,12]，左右子树的分割点是后面第一个比根节点大的数。接下来再分别对[5,1,7]和[10,12]做同样的操作，可以知道5是8的左子树根节点，1和7分别在其左右；而10是8的右子树根节点，12为右子树节点。很显然这是一个递归的过程，只要找到每个子树的根节点将其左右子树划分即可。

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 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