23天【代码随想录算法训练营34期】第六章 二叉树part09（● 669. 修剪二叉搜索树 ● 108.将有序数组转换为二叉搜索树 ● 538.把二叉搜索树转换为累加树 ● 总结篇）

669. 修剪二叉搜索树

# 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 trimBST(self, root: Optional[TreeNode], low: int, high: int) -> Optional[TreeNode]:
        if root is None:
            return None
        if root.val < low:
            root = self.trimBST(root.right, low, high)
            return root
        if root.val > high:
            root = self.trimBST(root.left, low, high)
            return root
        root.left = self.trimBST(root.left, low, high)
        root.right = self.trimBST(root.right, low, high)
        return root

108.将有序数组转换为二叉搜索树
需要再做一遍，左括号右括号还是搞不明白

# 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]:
        return self.build(nums, 0, len(nums)-1)

    def build(self, nums, beg, end):
        if beg > end:
            return None
        mid = beg + (end - beg) // 2
        root = TreeNode(nums[mid])
        root.left = self.build(nums, beg, mid-1)
        root.right = self.build(nums, mid+1, end)
        return root

538.把二叉搜索树转换为累加树

# 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 __init__(self):
        self.count = 0
    def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root is None:
            return None
        root.right = self.convertBST(root.right)
        self.count += root.val
        root.val = self.count
        root.left = self.convertBST(root.left)
        return root

总结篇