问题描述
可以在二叉搜索树中打印字符而不是整数吗?我正在尝试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?请帮我将 int 转换为字符串或字符。提前致谢!
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)
解决方法
我能想到的最直接的方法是为数字添加一个偏移量,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