代码随想录算法训练营day25 | 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

669. 修剪二叉搜索树

看题解做出来的

class Solution:
    def trimBST(self, root: Optional[TreeNode], low: int, high: int) -> Optional[TreeNode]:
        if not root:
            return None
        if root.val < low:
            return self.trimBST(root.right, low, high)
        if root.val > high:
            return self.trimBST(root.left, low, high)

        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

构造平衡搜索二叉树，从中间开始划分二叉树的左子树和右子树

class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
        return self.traversal(nums, 0, len(nums)-1)

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

538.把二叉搜索树转换为累加树

右中左遍历，得到从大到小的序列，累计求和

本题中pre可以初始化为0，就不需要判断了

class Solution:
    def __init__(self):
        self.pre = None

    def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        self.convertBST(root.right)
        if self.pre is not None:
            root.val += self.pre
        self.pre = root.val
        self.convertBST(root.left)
        return root