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

题目以下：

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]作一样的操做，能够知道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，只要找到每一个子树的根节点将其左右子树划分便可。

代码以下：

# 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