Question

我希望打印出[6,5,10

相反，我只获得10：

class BST:
  arr = []

  def __init__(self):
    self.height = 0
    self.key = 0
    self.left = None
    self.right = None

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

  def populate(self):
     print("populating")
     print(self.key)

     if (self.left != None):
       BST.arr = BST.arr + [self.left.populate()]

     if (self.right != None):
       BST.arr = BST.arr + [self.right.populate()]

     return self.key

m1 = BST()
m1.key = 10
m2 = BST()    
m2.key = 5
m1.left = m2

print(m1.left != None)

m3 = BST()
m3.key = 6
m2.left = m3

res = m1.populate()
print(res)

Answer 1

这是一个按顺序遍历的函数：

def inOrder(node, order):
    if node is not None:
        inOrder(node.left, order)
        order.append(node.key)
        inOrder(node.right, order)

你可以编写一个包装函数，它确实返回一个列表并像这样使用它：

def inOrder(root):
    order = []
    inOrder(root, order)
    return order

编辑：或者您可以使用以下一个函数执行此操作：

def inOrder(node):
    if node is None:
        return []

    return inOrder(node.left) + [node.key] + inOrder(node.right)

Answer 2

解决方法是更新populate方法：

class BST:
  arr = []

  def __init__(self):
    self.height = 0
    self.key = 0
    self.left = None
    self.right = None

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

  def populate(self):
    self.arr = []
    temp = self
    while temp.left != None or temp.right!=None:
        self.arr.append(temp.key)
        if temp.left!=None:
            temp = temp.left
        elif temp.right!=None:
            temp = temp.right
        if temp.left == None and temp.right==None:
            self.arr.append(temp.key)
    return self.arr
m1 = BST()
m1.key = 10
m2 = BST()    
m2.key = 5
m1.left = m2

print(m1.left != None)

m3 = BST()
m3.key = 6
m2.left = m3

res = m1.populate()
print(res)

输出：

True
[10, 5, 6]

按顺序打印BST节点 - 仅打印第一个节点？

2 个答案: