1 class Solution:
 2     def __init__(self):
 3         self.root = None
 4 
 5     def construct(self,order,root,i):
 6         if i==len(order):
 7             return None
 8         else:
 9             if self.root==None:
10                 self.root = TreeNode(order[i])
11                 self.construct(order,self.root,i+1)
12             else:
13                 if order[i]<root.val:
14                     if root.left==None:
15                         root.left = TreeNode(order[i])
16                         self.construct(order,self.root,i+1)
17                     else:
18                         self.construct(order,root.left,i)
19                 else:
20                     if root.right==None:
21                         root.right = TreeNode(order[i])
22                         self.construct(order,self.root,i+1)
23                     else:
24                         self.construct(order,root.right,i)
25 
26     def bstFromPreorder(self, preorder: 'List[int]') -> 'TreeNode':
27         self.construct(preorder,self.root,0)
28         return self.root

比较简单的题目,递归方式构造二叉搜索树。

posted on 2019-03-10 22:01  Sempron2800+  阅读(139)  评论(0编辑  收藏  举报