python之树的实现

一、前言

树的术语是生物学、家谱学和几何学术语的一个特殊混合体。它很好的表示了层级结构，它的两个主要特征：

每个项都有多个子节点。
除了叫做根（树最顶端的节点）的特殊项，所有其他的项都只有一个父节点。

普通的树和二叉树的递归定义：

普通的树------普通的树要么为空，要么包含一个有限的节点的集合T。有一个和所有其他节点不同的节点r，称为根，此外，集合T-{r}被划分为不相连的子集，每个子集都是一个普通的树。
二叉树------一个二叉树要么为空，要么包含一个根节点，外加一个左子树和一个右子树，每个子树都是一个二叉树。

二、二叉搜索树（Binary Search Tree,BST）

在一个BST中，给定节点的左子树中的节点要小于它，其右子树中的节点要大于给定的节点。

2.1 二叉搜索树接口中的方法

BST方法	作用
tree.isEmpty()	如果为空，返回True，否则返回False
tree.__len__()	返回树中元素的数目
tree.__iter__()	对树进行一次前向遍历
tree.__str__()	返回字符串，表示树的形状
tree.__contains__(item)	如果item在树中，返回True，否则返回False
tree.__add__(otherTree)	返回一个新树，内容为tree和otherTree的内容
tree.__eq__(anyObject)	若tree==otherTree，返回True，否则返回False
tree.clear()	让树变成空树
tree.add(item)	将item添加到树中适当的位置
tree.remove(item)	从树中删除item，先验条件：item在树中
tree.replace(item,newItem)	使用newItem替代树中匹配的项，并返回匹配的项
tree.inorder()	返回一个迭代器，对树执行中序遍历
tree.postorder()	返回一个迭代器，对树执行后序遍历
tree.levelorder()	返回一个迭代器，对树执行层序遍历

2.2代码实现

2.2.1 创建抽象集类，并保存为abstractcollection.py文件，具体代码见博客：https://blog.csdn.net/lzh_12345/article/details/79759960中的“二、利用链表实现栈”之2.1的代码。

2.2.2创建二叉搜索树节点类BSTNode，并保存为bstnode.py文件

class BSTNode(object):"""Represents a node for a linked binary search tree."""def __init__(self, data, left = None, right = None):self.data = dataself.left = leftself.right = right

2.2.3创建LinkedStack和LinkedQueue类，并分别保存为linkedstack.py和linkedqueue.py文件。

LinkedStack类的具体代码：见博客：https://blog.csdn.net/lzh_12345/article/details/79759960中的“二、利用链表实现栈”。

LinkedQueue类的具体代码：见博客：https://blog.csdn.net/lzh_12345/article/details/79762861中的“二、队列的链表实现”。

2.2.4 链表实现BST

from abstractcollection import AbstractCollection
from bstnode import BSTNode
from linkedstack import LinkedStack
from linkedqueue import LinkedQueueclass LinkedBST(AbstractCollection):"""An link-based binary search tree implementation."""def __init__(self, sourceCollection = None):"""Sets the initial state of self, which includes thecontents of sourceCollection, if it's present."""self._root = NoneAbstractCollection.__init__(self, sourceCollection)# Accessor methodsdef __str__(self):"""Returns a string representation with the tree rotated90 degrees counterclockwise."""def recurse(node, level):s = ""if node != None:s += recurse(node.right, level + 1)s += "| " * levels += str(node.data) + "\n"s += recurse(node.left, level + 1)return sreturn recurse(self._root, 0)def __iter__(self):"""Supports a preorder traversal on a view of self."""if not self.isEmpty():stack = LinkedStack()stack.push(self._root)while not stack.isEmpty():node = stack.pop()yield node.dataif node.right != None:stack.push(node.right)if node.left != None:stack.push(node.left)def preorder(self):"""Supports a preorder traversal on a view of self."""return Nonedef inorder(self):"""Supports an inorder traversal on a view of self."""lyst = list()def recurse(node):if node != None:recurse(node.left)lyst.append(node.data)recurse(node.right)recurse(self._root)return iter(lyst)def postorder(self):"""Supports a postorder traversal on a view of self."""return Nonedef levelorder(self):"""Supports a levelorder traversal on a view of self."""return Nonedef __contains__(self, item):"""Returns True if target is found or False otherwise."""return self.find(item) != Nonedef find(self, item):"""If item matches an item in self, returns thematched item, or None otherwise."""def recurse(node):if node is None:return Noneelif item == node.data:return node.dataelif item < node.data:return recurse(node.left)else:return recurse(node.right)return recurse(self._root)# Mutator methodsdef clear(self):"""Makes self become empty."""self._root = Noneself._size = 0def add(self, item):"""Adds item to the tree."""# Helper function to search for item's position def recurse(node):# New item is less, go left until spot is foundif item < node.data:if node.left == None:node.left = BSTNode(item)else:recurse(node.left)# New item is greater or equal, # go right until spot is foundelif node.right == None:node.right = BSTNode(item)else:recurse(node.right)# End of recurse# Tree is empty, so new item goes at the rootif self.isEmpty():self._root = BSTNode(item)# Otherwise, search for the item's spotelse:recurse(self._root)self._size += 1def remove(self, item):"""Precondition: item is in self.Raises: KeyError if item is not in self.postcondition: item is removed from self."""if not item in self:raise KeyError("Item not in tree.""")# Helper function to adjust placement of an itemdef liftMaxInLeftSubtreeToTop(top):# Replace top's datum with the maximum datum in the left subtree# Pre:  top has a left child# Post: the maximum node in top's left subtree#       has been removed# Post: top.data = maximum value in top's left subtreeparent = topcurrentNode = top.leftwhile not currentNode.right == None:parent = currentNodecurrentNode = currentNode.righttop.data = currentNode.dataif parent == top:top.left = currentNode.leftelse:parent.right = currentNode.left# Begin main part of the methodif self.isEmpty(): return None# Attempt to locate the node containing the itemitemRemoved = NonepreRoot = BSTNode(None)preRoot.left = self._rootparent = preRootdirection = 'L'currentNode = self._rootwhile not currentNode == None:if currentNode.data == item:itemRemoved = currentNode.databreakparent = currentNodeif currentNode.data > item:direction = 'L'currentNode = currentNode.leftelse:direction = 'R'currentNode = currentNode.right# Return None if the item is absentif itemRemoved == None: return None# The item is present, so remove its node# Case 1: The node has a left and a right child#         Replace the node's value with the maximum value in the#         left subtree#         Delete the maximium node in the left subtreeif not currentNode.left == None \and not currentNode.right == None:liftMaxInLeftSubtreeToTop(currentNode)else:# Case 2: The node has no left childif currentNode.left == None:newChild = currentNode.right# Case 3: The node has no right childelse:newChild = currentNode.left# Case 2 & 3: Tie the parent to the new childif direction == 'L':parent.left = newChildelse:parent.right = newChild# All cases: Reset the root (if it hasn't changed no harm done)#            Decrement the collection's size counter#            Return the itemself._size -= 1if self.isEmpty():self._root = Noneelse:self._root = preRoot.leftreturn itemRemoveddef replace(self, item, newItem):"""If item is in self, replaces it with newItem andreturns the old item, or returns None otherwise."""probe = self._rootwhile probe != None:if probe.data == item:oldData = probe.dataprobe.data = newItemreturn oldDataelif probe.data > item:probe = probe.leftelse:probe = probe.rightreturn None

2.2.5 测试文件

from linkedbst import LinkedBSTdef main():tree = LinkedBST()print("Adding D B A C F E G")tree.add("D")tree.add("B")tree.add("A")tree.add("C")tree.add("F")tree.add("E")tree.add("G")print("\nExpect True for A in tree: ", "A" in tree)print("\nString:\n" + str(tree))clone = LinkedBST(tree)print("\nClone:\n" + str(clone))print("Expect True for tree == clone: ", tree == clone)print("\nFor loop: ", end="")for item in tree:print(item, end=" ")print("\n\ninorder traversal: ", end="")for item in tree.inorder(): print(item, end = " ")print("\n\npreorder traversal: ", end="")for item in tree.preorder(): print(item, end = " ")print("\n\npostorder traversal: ", end="")for item in tree.postorder(): print(item, end = " ")print("\n\nlevelorder traversal: ", end="")for item in tree.levelorder(): print(item, end = " ")print("\n\nRemoving all items:", end = " ")for item in "ABCDEFG":print(tree.remove(item), end=" ")print("\n\nExpect 0: ", len(tree))tree = LinkedBST(range(1, 16))print("\nAdded 1..15:\n" + str(tree))lyst = list(range(1, 16))import randomrandom.shuffle(lyst)tree = LinkedBST(lyst)print("\nAdded ", lyst, "\n" + str(tree))if __name__ == "__main__":main()

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