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 = rightdef 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_10def 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 |
希望本文所述对大家Python程序设计有所帮助。
