用分支遞歸地打印樹

Question

我試圖在Python中遞歸地打印一棵樹。出於某種原因，縮進不起作用（也許我太累了，在這一點上看到一個明顯的缺陷）。下面是我的工作結構/類定義：

class Tree(object): 
    def __init__(self, data): 
    self.data = data 
    self.branches = [] 

class Branch(object): 
    def __init__(self, value): 
    self.label = value 
    self.node = None 

正如你看到的，每棵樹都有分公司，其中有一個標籤，並點到另一個樹（這是node值可以看到有）。以下是我正在試圖打印出樹：

def __str__(self): 
    return self.tree_string(0) 

    def tree_string(self, indent): 
    indentation = indent * " " 
    result = indentation + str(self.data) + "\n"; 
    for branch in self.branches: 
     result += indentation + branch.label + ": \n" + branch.node.tree_string(indent + 2) 
    return result 

這是給我：

4 
Somewhat: 
    Yes 
Fuller: 
    3 
Correct: 
    8 
Caribbean: 
    2 
Wrong: 
    Wrong 
Correct: 
    Correct 

Italian: 
    Wrong 
Burger: 
    Correct 

Wrong: 
    Wrong 

Nothing: 
    Wrong 

當它應該是給我像

4 
Somewhat: 
    Correct 
Fuller: 
    3 
    Correct: 
    8 
    Caribbean: 
     2 
     Wrong: 
     Wrong 
     Correct: 
     Correct 
    Italian: 
     Wrong 
    Burger: 
    Correct 
    Wrong: 
    Wrong 
Nothing: 
    Wrong 

是什麼導致我的代碼有這些額外的換行符，並沒有適當的縮進？

更新

很肯定的數據是確定的。下面是一個修改版本，顯示這是確定的：

def tree_string(self, indent): 
    indentation = indent * " " 
    result = str(self.data); 
    if len(self.branches) > 0: 
     result += "[" 
     for branch in self.branches: 
     result += branch.label + ":" + branch.node.tree_string(indent + 2) + " " 
     result += "]" 
    return result 

..這使輸出

4[Somewhat:Correct Fuller:3[Correct:8[Caribbean:2[No:No Correct:Correct ] Italian:Wrong Burger:Correct ] Wrong:Wrong ] Nothing:Wrong ]

然而，縮進值，是由於某種原因，始終爲0或2

Answer 1

外貌像它應該對我工作：

class Tree(object): 
    def __init__(self, data): 
    self.data = data 
    self.branches = [] 
    def __str__(self): 
    return self.tree_string(0) 

    def tree_string(self, indent): 
    indentation = indent * " " 
    result = indentation + str(self.data) + "\n"; 
    for branch in self.branches: 
     result += indentation + branch.label + ": \n" + branch.node.tree_string(indent + 2) 
    return result 

class Branch(object): 
    def __init__(self, value): 
    self.label = value 
    self.node = None 

tree = Tree(4) 
b1 = Branch('Somewhat') 
b1.node = Tree('Yes') 
b2 = Branch('Fuller') 
b2.node = Tree(3) 
tree.branches = [b1, b2] 
b3 = Branch('Correct') 
b3.node = Tree(8) 
b2.node.branches = [b3] 
print(tree) 

yie lds

4 
Somewhat: 
    Yes 
Fuller: 
    3 
    Correct: 
    8