一个简单的二叉树实现，简单二叉树实现,[Python]代码#!

一个简单的二叉树实现，简单二叉树实现,[Python]代码#!

[Python]代码

#!/usr/bin/env python# -*- coding: utf-8 -*-import osclass Node(object):    """docstring for Node"""    def __init__(self, v = None, left = None, right=None, parent=None):        self.value = v        self.left = left        self.right = right        self.parent = parentclass BTree(object):    """docstring for BtTee """    def __init__(self):        self.root = None        self.size = 0    def insert(self, node):        n = self.root        if n == None:            self.root = node            return        while True:            if node.value <= n.value:                if n.left == None:                    node.parent = n                    n.left = node                    break                else:                    n = n.left            if node.value > n.value:                if n.right == None:                    n.parent = n                    n.right = node                    break                else:                    n = n.right    def find(self, v):        n = self.root        while True:            if n == None:                return None            if v == n.value:                return n            if v < n.value:                n = n.left                continue            if v > n.value:                n = n.right    def find_successor(node):        '''查找后继结点'''        assert node != None and node.right != None        n = node.right        while n.left != None:            n = n.left        return n    def delete(self, v):        n = self.find(v)        print "delete:",n.value        del_parent = n.parent        if del_parent == None:            self.root = None;            return        if n != None:            if n.left != None and n.right != None:                succ_node = find_successor(n)                parent = succ_node.parent                if succ_node == parent.left:                    #if succ_node is left sub tree                    parent.left = None                if succ_node == parent.right:                    #if succ_node is right sub tree                    parent.right = None                if del_parent.left == n:                    del_parent.left = succ_node                if del_parent.right == n:                    del_parent.right = succ_node                succ_node.parent = n.parent                succ_node.left = n.left                succ_node.right = n.right                del n            elif n.left != None or n.right != None:                if n.left != None:                    node = n.left                else:                    node = n.right                node.parent = n.parent                if del_parent.left == n:                    del_parent.left = node                if del_parent.right == n:                    del_parent.right = node                del n            else:                if del_parent.left == n:                    del_parent.left = None                if del_parent.right == n:                    del_parent.right = None    def tranverse(self):        def pnode(node):            if node == None:                return            if node.left != None:                pnode(node.left)            print node.value            if node.right != None:                pnode(node.right)        pnode(self.root)def getopts():    import optparse, locale    parser = optparse.OptionParser()    parser.add_option("-i", "--input", dest="input", help=u"help name", metavar="INPUT")    (options, args) = parser.parse_args()    #print options.input    return (options.input)if __name__ == '__main__':    al = [23, 45, 67, 12, 78,90, 11, 33, 55, 66, 89, 88 ,5,6,7,8,9,0,1,2,678]    bt = BTree()    for x in al :        bt.insert(Node(x))    bt.delete(12)    bt.tranverse()    n = bt.find(12)    if n != None:        print "find valud:",n.value