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55 lines (48 loc) · 1.36 KB
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#!/usr/bin/env python
# @Time : 2018/5/13 下午1:07
# @Author : cancan
# @File : method_1.py
# @Function : 将有序数组转换为二叉搜索树
"""
Question:
将一个按照升序排列的有序数组,转换为一棵高度平衡二叉搜索树。
本题中,一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。
Example:
给定有序数组: [-10,-3,0,5,9],
一个可能的答案是:[0,-3,9,-10,null,5],它可以表示下面这个高度平衡二叉搜索树:
0
/ \
-3 9
/ /
-10 5
"""
# 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 sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
if not nums:
return None
def f(node, n):
l = len(n)
if l == 1:
node.val = n[0]
return
d = l // 2
node.val = n[d]
if n[:d]:
node.left = TreeNode(0)
f(node.left, n[:d])
if n[d + 1:]:
node.right = TreeNode(0)
f(node.right, n[d + 1:])
r = TreeNode(0)
f(r, nums)
return r