时间:2021-07-01 10:21:17 帮助过:98人阅读
- # 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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