本文实例讲述了Python二叉搜索树与双向链表实现方法。分享给大家供大家参考，具体如下：
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# encoding=utf8
'''
题目：输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。
要求不能创建任何新的结点，只能调整树中结点指针的指向。
'''
class BinaryTreeNode():
def __init__(self, value, left = None, right = None):
self.value = value
self.left = left
self.right = right
def create_a_tree():
node_4 = BinaryTreeNode(4)
node_8 = BinaryTreeNode(8)
node_6 = BinaryTreeNode(6, node_4, node_8)
node_12 = BinaryTreeNode(12)
node_16 = BinaryTreeNode(16)
node_14 = BinaryTreeNode(14, node_12, node_16)
node_10 = BinaryTreeNode(10, node_6, node_14)
return node_10
def print_a_tree(root):
if root is None:return
print_a_tree(root.left)
print root.value, ' ',
print_a_tree(root.right)
def print_a_linked_list(head):
print 'linked_list:'
while head is not None:
print head.value, ' ',
head = head.right
print ''
def create_linked_list(root):
'''构造树的双向链表，返回这个双向链表的最左结点和最右结点的指针'''
if root is None:
return (None, None)
# 递归构造出左子树的双向链表
(l_1, r_1) = create_linked_list(root.left)
left_most = l_1 if l_1 is not None else root
(l_2, r_2) = create_linked_list(root.right)
right_most = r_2 if r_2 is not None else root
# 将整理好的左右子树和root连接起来
root.left = r_1
if r_1 is not None:r_1.right = root
root.right = l_2
if l_2 is not None:l_2.left = root
# 由于是双向链表，返回给上层最左边的结点和最右边的结点指针
return (left_most, right_most)
if __name__ == '__main__':
tree_1 = create_a_tree()
print_a_tree(tree_1)
(left_most, right_most) = create_linked_list(tree_1)
print_a_linked_list(left_most)
pass
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希望本文所述对大家Python程序设计有所帮助。