返回二叉树的高度，返回比实际高度大一的函数

Question

[只有高度函数是我编写的，因此我对其余的代码没有任何控制。我不知道怎么了请帮助，在此先感谢

class Node:
    def __init__(self, info): 
        self.info = info  
        self.left = None  
        self.right = None 
        self.level = None 

    def __str__(self):
        return str(self.info) 

class BinarySearchTree:
    def __init__(self): 
        self.root = None

    def create(self, val):  
        if self.root == None:
            self.root = Node(val)
        else:
            current = self.root

            while True:
                if val < current.info:
                    if current.left:
                        current = current.left
                    else:
                        current.left = Node(val)
                        break
                elif val > current.info:
                    if current.right:
                        current = current.right
                    else:
                        current.right = Node(val)
                        break
                else:
                    break

# Enter your code here. Read input from STDIN. Print output to STDOUT
'''
class Node:
      def __init__(self,info): 
          self.info = info  
          self.left = None  
          self.right = None 


       // this is a node of the tree , which contains info as data, left , right
'''
def height(root):
    if root == None:
        return 0
    else:
        leftheight= 1+height(root.left)
        rightheight= 1+height(root.right)
    if leftheight>rightheight:
        return leftheight
    else:
        return rightheight




tree = BinarySearchTree()
t = int(input())

arr = list(map(int, input().split()))

for i in range(t):
    tree.create(arr[i])

print(height(tree.root))

输出总是比实际高度小一，例如，如果树的高度为3，则返回4

Answer 1

尝试一下：

def height(root):
    if root is None:
        return 0
    else:
        return 1 + max(height(root.left), height(root.right))