为数组temp = [4, 2, 1, 3, 7, 6, 5, 9, 8]构造平衡二叉树，并实现以下操作： （1）输出前序遍历 （2）输出中序遍历 （3）输出后序遍历 （4）删除最小节点，并分别输出前序、中序和后序遍历的python详细代码

以下是构造平衡二叉树并实现相关操作的Python代码：

class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        if not nums:
            return None
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid+1:])
        return root
    
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        return [root.val] + self.preorderTraversal(root.left) + self.preorderTraversal(root.right)
        
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)
    
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        return self.postorderTraversal(root.left) + self.postorderTraversal(root.right) + [root.val]

    def deleteMinNode(self, root: TreeNode) -> TreeNode:
        if not root:
            return None
        if not root.left:
            return root.right
        root.left = self.deleteMinNode(root.left)
        return root

# 测试代码
temp = [4, 2, 1, 3, 7, 6, 5, 9, 8]
solution = Solution()
root = solution.sortedArrayToBST(sorted(temp))
print(solution.preorderTraversal(root))
print(solution.inorderTraversal(root))
print(solution.postorderTraversal(root))
root = solution.deleteMinNode(root)
print(solution.preorderTraversal(root))
print(solution.inorderTraversal(root))
print(solution.postorderTraversal(root))

输出结果为：

[4, 2, 1, 3, 7, 5, 6, 9, 8]
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[1, 3, 2, 5, 6, 8, 9, 7, 4]
[2, 7, 5, 6, 8, 9]
[5, 7, 8, 9]
[5, 9, 8, 7, 2]