【发布时间】：2017-09-16 09:15:16
【问题描述】：
我正在准备考试，但我不知道如何回答这个问题。
创建一个返回二叉树后遍历的链表的函数。您不能使用任何容器预先存储数据，并且它必须是递归的。
所以我不能进行post遍历，把它放在一个列表中，然后遍历它来制作一个链表。
还有另一个问题说要做同样的事情，但要进行中序遍历，这就是解决方案。
有两种解决方案。一个我不明白的，一个清晰的版本
这是主要的解决方案
def inorder(root: BTNode) -> LLNode:
"""Return the first node in a linked list that contains every value from the
binary tree rooted at root, listed according to an inorder traversal.
>>> b = BTNode(1, BTNode(2), BTNode(3))
>>> repr(inorder(b))
'LLNode(2, LLNode(1, LLNode(3)))'
>>> b2 = BTNode(4, BTNode(5))
>>> b3 = BTNode(7, b, b2)
>>> str(inorder(b3))
'2 -> 1 -> 3 -> 7 -> 5 -> 4'
>>> # from the handout...
>>> left = BTNode('B', None, BTNode('D', BTNode('G')))
>>> right = BTNode('C', BTNode('E'), BTNode('F'))
>>> root = BTNode('A', left, right)
>>> str(inorder(root))
'B -> G -> D -> A -> E -> C -> F'
"""
return _inorder(root)[0]
def _inorder(root: BTNode) -> (LLNode, LLNode):
"""Return the first and last nodes in a linked list that contains every
value from the binary tree rooted at root, listed according to an inorder
traversal.
"""
if root:
head_left, tail_left = _inorder(root.left)
head_right, tail_right = _inorder(root.right)
node_root = LLNode(root.item, head_right)
if tail_left:
tail_left.link = node_root
return head_left or node_root, tail_right or node_root
else:
return None, None
明确的解决方案
def inorder(root: BTNode) -> LLNode:
"""Return the first node in a linked list that contains every value from the
binary tree rooted at root, listed according to an inorder traversal.
>>> b = BTNode(1, BTNode(2), BTNode(3))
>>> repr(inorder(b))
'LLNode(2, LLNode(1, LLNode(3)))'
>>> b2 = BTNode(4, BTNode(5))
>>> b3 = BTNode(7, b, b2)
>>> str(inorder(b3))
'2 -> 1 -> 3 -> 7 -> 5 -> 4'
>>> # from the handout...
>>> left = BTNode('B', None, BTNode('D', BTNode('G')))
>>> right = BTNode('C', BTNode('E'), BTNode('F'))
>>> root = BTNode('A', left, right)
>>> str(inorder(root))
'B -> G -> D -> A -> E -> C -> F'
"""
return _inorder(root)[0] # what must this first item represent?
def _inorder(root: BTNode) -> (LLNode, LLNode): # what are these 1st and 2nd things?
"""Return the first and last nodes in a linked list that contains every
value from the binary tree rooted at root, listed according to an inorder
traversal.
>>> left = BTNode('B', None, BTNode('D', BTNode('G')))
>>> right = BTNode('C', BTNode('E'), BTNode('F'))
>>> root = BTNode('A', left, right)
>>> str(inorder(root))
'B -> G -> D -> A -> E -> C -> F'
"""
if not root:
return None, None
else:
# Start off by making a new node of our item, with None for a link
# Obviously we will need to replace that None if we have a right branch
new_node = LLNode(root.item)
# Recursive call on right branch gives us its head and tail
right_head, right_tail = _inorder(root.right)
# The link on our new node should be the right head, even if it's None
new_node.link = right_head
# Ultimately the tail for this whole node will be the rightmost tail
# If there is no right side, though, this node is the rightmost tail
if not right_tail:
right_tail = new_node
# Recursive call on left branch gives us its head and tail
left_head, left_tail = _inorder(root.left)
# If there is a left tail, we should string our current node to the end
if left_tail:
left_tail.link = new_node
# Ultimately the head for this whole node will be the leftmost head
# If there is no left head, though, this node is the leftmost head
if not left_head:
left_head = new_node
# Return the leftmost head and the rightmost tail
return left_head, right_tail
【问题讨论】：
标签： algorithm python-3.x recursion linked-list binary-tree