Make a little progress every day , It's already great , Insist on , Don't be too tired , Refuse to roll in , Start with practice every day , Ten minutes a day , Live happily all your life ！ The epidemic is still repeated , Everyone wear masks ~ Continue to continue , Come on , Today and Brother cheshen Work together to improve your Python Programming and Interview ability Well , brush ladder on high buildings ~
Put on what I took Photo Well ！ I bought a hundred courses of tea in the evening , Reward yourself with a wave of , ha-ha
Recommend a song every day ： A north ——Jay Chou
The following is my Ladder integral rule ：
At least one question a day ： An integral +10 branch
If you do one more question （ Or one more way to answer ）, Then the points of the day +20 branch （+10+10）
If you do more than three , Start with question 3 +20 branch （ Such as ： If you do three questions, you will score -10+10+20=40; If you do four questions, you will score –10+10+20+20=60）Initial classification 100 branch
If you haven't done the problem for one day , Then deduct points -10 branch （ Saturday 、 Except Sunday ： rest ）
insist ！！！
Give you an array of integers nums , Where elements have been Ascending array , Please turn it into a Highly balanced Binary search tree .
Highly balanced A binary tree is a tree that satisfies 「 The absolute value of the height difference between the left and right subtrees of each node does not exceed 1 」 Two fork tree .
Example 1：
Input ：nums = [-10,-3,0,5,9]
Output ：[0,-3,9,-10,null,5]
explain ：[0,-10,5,null,-3,null,9] Will also be considered the right answer ：
Example 2：
Input ：nums = [1,3]
Output ：[3,1]
explain ：[1,3] and [3,1] They're all highly balanced binary search trees .
analysis ：
What we need to do is transform it into a highly balanced binary search tree , The title says our array nums It's the one that has finished arranging the serial number , We can use the idea of recursion , The middle value in the array will be taken out every time , Then take the previous number as the left node , The latter number is used as the node of the right subtree , In this way, the allocation of binary tree is completed . We can use slices to recursively take values , Then fill in the sub tree species . Today's question is not very difficult , There is also a deep search to try .
The big man's picture is simply the voice of his heart ：
class Solution: def sortedArrayToBST(self, nums: List[int]) -> TreeNode: n = len(nums) if n == 0: return None mid = n // 2 # Plate division , Median root = TreeNode(nums[mid]) root.left = Solution.sortedArrayToBST(self, nums[:mid]) root.right = Solution.sortedArrayToBST(self, nums[mid+1:]) return root
Binary tree , Come here , Also finished painting , " ！
I'm a little free now , Today was too tired , Finally the weekend , Take a rest , incorrect , There has to be a meeting tomorrow , ah ！！！
Tomorrow I'll sleep until noon ！！！
