【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