计算二叉搜索树中不同类型的节点

Question

我正在尝试编写一个计算不同类型节点的递归函数：

• 具有零子节点的节点数
• 一个孩子的节点数

• 有两个孩子的节点数

  def _node_counts_aux(self,node):
  
      zero = 0
      one = 0
      two = 0
  
      if node is not None:     
          #print("==========")
          #print(node)
          #print(node._left)
          #print(node._right)
          #print()
  
          # Two children case
          if node._left is not None and node._right is not None:
              #print("Two")
              two = self._node_counts_aux(node._left)[2] + self._node_counts_aux(node._right)[2] + 1
  
          # One child case
          elif node._left is None and node._right is not None:
              #print("One")
              one = self._node_counts_aux(node._right)[1] + 1
  
          elif node._right is None and node._left is not None:
              #print("One")
              one = self._node_counts_aux(node._left)[1] + 1
  
          # Zero children case
          elif node._left is None and node._right is None:
              #print("Zero")
              zero = self._node_counts_aux(node._left)[0] + self._node_counts_aux(node._right)[0] + 1
  
      return zero, one, two
  

我相信if 语句结构涵盖了所有情况，但是当我使用这个 bst 运行我的程序时，我得到了这个结果：

Zero children: 0
One child:     0
Two children:  4

我使用 print 语句测试了我的程序，并确定我的递归调用存在问题。然而，问题是，一切（据我所知）都是正确的。有人可以指出我的错误吗？提前致谢

Answer 1

您的函数实际上返回后续完整节点的数量。您应该保留每次呼叫的计数器信息。

def _node_counts_aux(self,node):
    # 'n[i]' is the total number of nodes with 'i' sons
    n = [0,0,0]
    sons = 0

    #one call for left 
    if node._left:
        n = [sum(x) for x in zip(n, self._node_counts_aux(node._left))]
        sons += 1

    #one call for right 
    if node._right:
        n = [sum(x) for x in zip(n, self._node_counts_aux(node._right))]
        sons += 1

    #count this node in the relevant counter 
    n[sons] += 1

    return n

更好（代码重用）：

def _node_counts_aux(self,node):
    # 'n[i]' is the total number of nodes with 'i' sons
    n = [0,0,0]
    sons = 0

    for son_node in [node._left, node._right]:
        if son_node:
            n = [sum(x) for x in zip(n, self._node_counts_aux(son_node))]
            sons += 1

    #count this node in the relevant counter 
    n[sons] += 1

    return n