Question

我编写了这段代码，当我使用print时，我看到了叶子。但是，该函数的最终返回值是None，而不是叶子的总和，在此示例中，叶子的总和应为7。我很高兴知道这里出了什么问题。谢谢！

class Node:
    def __init__(self, val=None):
        self.left = None
        self.right = None
        self.val = val


def sum_leafs(tree):
    if tree is None:
        return 0

    if tree.right and tree.left:
        sum_leafs(tree.right)
        sum_leafs(tree.left)

    elif tree.right or tree.left:
        if tree.right:
            sum_leafs(tree.right)
        elif tree.left:
            sum_leafs(tree.left)

    elif tree.right is None and tree.left is None:
        return sum_leafs(tree.left) + 1


node = Node(10)
node.right = Node(2)
node.left = Node(11)
node.left.right = Node(5)

print(sum_leafs(node))

Answer 1

在对分支（左/右）求和时，您忘记添加+，并且也忘记了访问val，这对于整个工作至关重要。

进一步，逻辑可以简化：

def sum_leafs(tree):
    if tree is None:
        return 0

    if not tree.right and not tree.left:
        return tree.val

    return sum_leafs(tree.right) + sum_leafs(tree.left)

Answer 2

您没有将总和相加或归还。这也可以通过类中的方法来完成：

class Node:
    def __init__(self, val=None):
        self.left = None
        self.right = None
        self.val = val

    def sum(self):
        s = 0
        if self.left is not None:
            s += self.left.sum()
        if self.right is not None:
            s += self.right.sum()
        return self.val + s


node = Node(10)
node.right = Node(2)
node.left = Node(11)
node.left.right = Node(5)

print(node.sum())

返回：

Answer 3

您没有正确返回计算出的叶子和。试试这个：

class Node:
    def __init__(self, val=None):
        self.left = None
        self.right = None
        self.val = val


def sum_leafs(tree):
    if tree is None:
        return 0

    elif tree.right and tree.left:
        return sum_leafs(tree.right) + sum_leafs(tree.left)

    elif tree.right or tree.left:
        if tree.right:
            return sum_leafs(tree.right)
        elif tree.left:
            return sum_leafs(tree.left)

    elif tree.right is None and tree.left is None:
        return tree.val

node = Node(10)
node.right = Node(2)
node.left = Node(11)
node.left.right = Node(5)

print(sum_leafs(node))

Answer 4

首先，我将更新您的Node界面，以便在创建节点时可以设置left和right分支-

class Node:
  def __init__(self, val=None, left=None, right=None):
    self.left = left
    self.right = right
    self.val = val

这使我们可以更符合人体工程学地创建发束，例如-

t = Node(10, Node(11, None, Node(5)), Node(2))

现在，我们编写一个通用遍历过程。这使我们能够将1）树的遍历与2）我们要在每个树元素上执行的预期操作分开-

def traverse(tree):
  if tree is None:
    return
  else:
    yield tree.val
    yield from traverse(tree.left)
    yield from traverse(tree.right)

现在sum_leafs可以是一个简单程序。它不再需要关注树的遍历，而可以专注于将值的线性序列转换为所需的总和-

def sum_leafs(tree):
  r = 0
  for val in traverse(tree):
    r += val
  return r

print(sum_leafs(t))
# 28

或者，我们可以编写一个search函数来查找传递谓词的第一个值，而不是对这些值求和-

def search(test, tree):
  for val in traverse(tree):
    if test(val):
      return val

print(search(lambda x: x < 10, t))
# 5

print(search(lambda x: x > 99, t))
# None

或者，我们可以简单地将每个值收集到一个列表中-

print(list(traverse(t)))
# [ 10, 11, 5, 2 ]

如您所见，从依赖于我们的树的每个函数中删除遍历逻辑将是巨大的帮助。

如果您不喜欢生成器，则可以编写急切版本的traverse，该版本总是返回list。现在的区别是无法部分遍历树。请注意，该程序与生成器版本具有相似之处-

def traverse(tree):
  if tree is None:
    return [] # <-- empty
  else:
    return \
      [ tree.val
      , *traverse(tree.left)  # <-- yield from
      , *traverse(tree.right) # <-- yield from
      ]

print(traverse(t))
# [ 10, 11, 5, 2 ]

二叉树叶子值的总和

4 个答案: