算法--二叉搜索树的python实现

Posted 2021-02-27 Knight66666

二叉查找树（Binary Search Tree），也称为二叉搜索树、有序二叉树或排序二叉树，是指一棵空树或者具有下列性质的二叉树：

1. 若任意节点的左子树不空，则左子树上所有节点的值均小于它的根节点的值；
2. 若任意节点的右子树不空，则右子树上所有节点的值均大于它的根节点的值；
3. 任意节点的左、右子树也分别为二叉查找树；
4. 没有键值相等的节点。

通常采取二叉链表作为二叉查找树的存储结构（使用队列存储）。

# encoding=utf-8
import queue
 
class TreeNode(object):
    def __init__(self,val):
        self.value = val
        self.left = None
        self.right = None
        self.father = None

# 定义二叉搜索树的各种功能方法      
class BST(object):
    def __init__(self,nodeList):
        self.root = None
        for node in nodeList:
            self.bfs_insert(node)
        #self.bfs_insert(node)中的bfs_insert方法在后面实现。放在构造函数中的目的是将一个列表生成一个二叉搜索树列表。
    
    #实现插入功能
    #第一步：将根节点（self.root）设置成当前位置节点（cur），即从根节点开始遍历。根节点的父节点（father）初始设置为None。 
    def bfs_insert(self,node):
        father = None
        cur = self.root
    #  第二步：查找可以插入的位置

 

    # 1. 当前位置节点（cur）的值与待插入节点（node）的值相等，返回-1（代表无法进行插入操作）。并将父节点（father）值修改为当前位置节点（cur），代表该节点已经遍历完成。
        while cur != None:
            if cur.value == node.value:
                return -1
            father = cur
            if node.value < cur.value:
                cur = cur.left
            else:
                cur = cur.right
        node.father = father
        if father == None:
            self.root = node
        elif node.value < father.value:
            father.left = node
        else:
            father.right = node
 
    def bfs(self):
        if self.root == None:
            return None
        retList = []
        q = queue.Queue()
        q.put(self.root)
        while q.empty() is not True:
            node = q.get()
            retList.append(node.value)
            if node.left != None:
                q.put(node.left)
            if node.right != None:
                q.put(node.right)
        return retList
 
    def bfs_search(self,value):
        cur = self.root
        while cur != None:
            if cur.value == value:
                return cur
            elif cur.value < value:
                cur = cur.right
            else:
                cur = cur.left
        return None
 
    def bfs_delete(self,node):
        father = node.father
        if node.left == None:
            if father == None:
                self.root = node.right
                if node.right != None:
                    node.right.father = None
            elif father.left == node:
                father.left = node.right
                if node.right != None:
                    node.right.father = father
            else:
                father.right = node.right
                if node.right != None:
                    node.right.father = father
            return ‘delete successfully‘
        tmpNode = node.left
        while tmpNode.right != None:
            tmpNode = tmpNode.right
 
        tmpNode.right = node.right
        if node.right != None:
            node.right.father = tmpNode
 
        if father == None:
            self.root = node.left
            node.left.father = None
        elif father.left == node:
            father.left = node.left
            node.left.father = father
        else:
            father.right = node.left
            node.left.father = father
        node = None
        return ‘delete successfully‘

    
    
if __name__ == ‘__main__‘:
    varList = [24,34,5,4,8,23,45,35,28,6,29]
    nodeList = [TreeNode(var) for var in varList]
    bst = BST(nodeList)
    print (bst.bfs())
    
    node = bst.bfs_search(34)
    bst.bfs_delete(node)
    print (bst.bfs())
    node1 = TreeNode(1)
    bst.bfs_insert(node1)
    print (bst.bfs())