題目:

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

給定一個sorted list,將它轉變成高度平衡二元樹(height-balanced binary tree)
高度平衡二元樹定義:所有兩子樹高度最多相差1的binary tree
ex:input:[-10,-3,0,5,9]=>output: [0,-3,9,-10,null,5]or[0,-10,5,null,-3,null,9]

按照題目給的範例,可以看出規律是取mid當root,分開的兩串陣列再取mid當左右節點,以此類推
我們按照這樣的理念實作程式

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
        if len(nums)==0: #當沒東西時回傳None,不開新節點
            return None
            
        if len(nums)==1: #當值只有一個時,就開此值的節點就好
            return TreeNode(nums[0])
            
        mid=len(nums)//2
        root=TreeNode(nums[mid])
        root.right=self.sortedArrayToBST(nums[mid+1:len(nums)])
        root.left=self.sortedArrayToBST(nums[0:mid])
        return root

先開一個值為nums[mid]的節點
mid兩邊的list用一樣的手法遞迴不斷往下開左右節點,直到list被切割到長度只剩0或1為止
最後執行時間63ms(faster than 93.48%)

那我們下題見