二进制搜索树插入字符

Question

是否可以在二进制搜索树中打印字符而不是整数？我正在试着打印字母而不是数字。有没有办法把1、2、3、4、5、6和7变成a、b、c、d、e、f和g？请帮我把int转换成字符串或字符。提前谢谢！

class Node:
def __init__(self, key):
self.left = None
self.right = None
self.val = key
def printInorder(root):
if root:
printInorder(root.left)
print(root.val),
printInorder(root.right)

def printPostorder(root):
if root:
printPostorder(root.left)
printPostorder(root.right)
print(root.val),
def printPreorder(root):
if root:
print(root.val),
printPreorder(root.left)
printPreorder(root.right)
root = Node(1)
root.right = Node(2)
root.right.right = Node(3)
root.right.right.right = Node(4)
root.right.right.right.right = Node(5)
root.right.right.right.right.right = Node(6)
root.right.right.right.right.right.right = Node(7)
print ("Preorder")
printPreorder(root) 
print ("In-order")
printInorder(root)
print ("Post-order")
printPostorder(root)

Answer 1

我能想到的最直接的方法是在数字上添加偏移量，然后将其转换为字符：

def print_as_letter(num):
ascii_a = 97
assert num < 27, "not enough letters in the alphabet"
letter = chr(num+ascii_a-1) # -1 since you start your nodes at 1
print(letter)
class Node:
def __init__(self, key):
self.left = None
self.right = None
self.val = key
def printInorder(root):
if root:
printInorder(root.left)
print_as_letter(root.val)
printInorder(root.right)

def printPostorder(root):
if root:
printPostorder(root.left)
printPostorder(root.right)
print_as_letter(root.val)
def printPreorder(root):
if root:
print_as_letter(root.val)
printPreorder(root.left)
printPreorder(root.right)
root = Node(1)
root.right = Node(2)
root.right.right = Node(3)
root.right.right.right = Node(4)
root.right.right.right.right = Node(5)
root.right.right.right.right.right = Node(6)
root.right.right.right.right.right.right = Node(7)
print ("Preorder")
printPreorder(root) 
print ("In-order")
printInorder(root)
print ("Post-order")
printPostorder(root)

输出：

Preorder
a
b
c
d
e
f
g
In-order
a
b
c
d
e
f
g
Post-order
g
f
e
d
c
b
a