python实战--数据结构二叉树

此文将讲述如何用python实战解决二叉树实验

前面已经讲述了python语言的基本用法，现在让我们实战一下具体明确python的用法

点击我进入python速成笔记

先看一下最终效果图：

首先我们要定义二叉树结点的一个类，在python中定义二叉树结点代码如下:

#二叉链表
class BiTree:
    def __init__(self, elementType=None, lchild=None, rchild=None):
        self.elementType = elementType
        self.lchild = lchild
        self.rchild = rchild

其次初始化二叉树头结点的代码如下：

#初始化二叉树新建头结点
def initTree(s):
    temp=BiTree()
    ptree=BiTree("A",temp,temp)
    if s[2]==0:
        ptree.lchild=None
    if s[4]==0:
        ptree.rchild=None
    return ptree,temp

二叉树中经常需要访问子节点，那么子节点的代码：

#寻找下一个结点
def getNext(tree,temp):
    if(tree.lchild ==temp ):
        return tree
    if (tree.lchild!=None):
        if(getNext(tree.lchild,temp)!=None):
            return getNext(tree.lchild,temp)
    if (tree.rchild==temp):
        return tree
    if(tree.rchild !=None):
        if (getNext(tree.rchild,temp) != None):
            return getNext(tree.rchild,temp)
    return None

有了头结点也能找到子节点，那么绘图二叉树的代码如下：

绘图需要传入图形窗口对象，根节点，初始根节点x，y坐标以及深度1

#画出二叉树图
def drawroot(graph,root,x,y,deep):
    # 画椭圆
    oval = Oval(Point(x-20, y-20), Point(x+20, y+20))
    oval.setFill('blue')  # 填充颜色
    oval.draw(graph)
    # 显示文字
    message = Text(Point(x, y), root.elementType)
    message.draw(graph)
    if(root.lchild!=None):
        # 画线
        line = Line(Point(x - 10, y + 10), Point(x - 30*deep, y + 60))
        line.draw(graph)
        drawroot(graph,root.lchild,x-30*deep,y+60,deep-1)
    if(root.rchild!=None):
        # 画线
        line = Line(Point(x+10, y + 10), Point(x + 30*deep, y + 60))
        line.draw(graph)
        drawroot(graph, root.rchild, x + 30*deep, y + 60,deep-1)

如何快速生成二叉树需要动用文件，根据文件链接子树代码：

读入根节点和一个临时结点，根节点名字和是否有左右子树开始链接

#根据文件建立二叉链表
def getTree(tree,temp,name,left,right):
    child=BiTree(name,temp,temp)
    if(not left):
        child.lchild=None
    if(not right):
        child.rchild=None
    if(tree.lchild==temp):
        tree.lchild=child
    elif(tree.rchild==temp):
        tree.rchild=child

具体完整读取代码完全链接子树代码如下：

input = open('bt31.txt', 'r')
s = []
try:
    for line in input:
        s.append(line)
finally:
    input.close()
ptree, temp = initTree(s[0])
for i in range(len(s)):
    if i != 0:
        getTree(getNext(ptree, temp), temp, s[i][0], eval(s[i][2]), eval(s[i][4]))

然后生成图形窗口绘制上面二叉树代码为：

gragh = GraphWin('CSSA', 1200, 700)
    drawroot(gragh, ptree, 500, 20, 4)

效果如下：

以上便是二叉树生成的前提操作，后面代码即为数据结构实验五二叉树1-10题必做题源代码以及12，13，15选做题源码：

#三种二叉遍历
def DLR(tree):
    order=[]
    if(tree!=None):
        order.append(tree.elementType)
        if(tree.lchild!=None):
            order=order+DLR(tree.lchild)
        if(tree.rchild!=None):
            order=order+DLR(tree.rchild)
    return order
def LDR(tree):
    order=[]
    if (tree != None):
        if (tree.lchild != None):
            order = order + LDR(tree.lchild)
        order.append(tree.elementType)
        if (tree.rchild != None):
            order = order + LDR(tree.rchild)
    return order
def LRD(tree):
    order=[]
    if (tree != None):
        if (tree.lchild != None):
            order = order + LRD(tree.lchild)
        if (tree.rchild != None):
            order = order + LRD(tree.rchild)
        order.append(tree.elementType)
    return order
#打印题目一信息
def exp1(gragh,ptree):
    str=""
    str += "先序遍历："
    for i in range(len(DLR(ptree))):
        str+=DLR(ptree)[i]
    str +="\n"
    str += "中序遍历："
    for i in range(len(LDR(ptree))):
        str +=LDR(ptree)[i]
    str +="\n"
    str += "后序遍历："
    for i in range(len(LRD(ptree))):
        str +=LRD(ptree)[i]
    str +="\n"
    message = Text(Point(200, 400), "1.打印出二叉树的三种遍历序\n"+str)
    message.draw(gragh)
#遍历结点层次
def DLRDeep(tree,deep):
    order=""
    if(tree!=None):
        order=order+tree.elementType+"        "+repr(deep)+"\n"
        if(tree.lchild!=None):
            order=order+DLRDeep(tree.lchild,deep+1)
        if(tree.rchild!=None):
            order=order+DLRDeep(tree.rchild,deep+1)
    return order
def exp2(gragh,ptree):
    str=DLRDeep(ptree,1)
    message = Text(Point(1000, 300), "2.输出各结点的层次\n"+str)
    message.draw(gragh)
#查找树的高度
def height(tree):
    h=0
    if(tree!=None):
        left=height(tree.lchild)
        right=height(tree.rchild)
        if(left>right):
            h=left+1
        else :
            h=right+1
    return h
def exp3(gragh,ptree):
    str="3.该二叉树高度为"+repr(height(ptree))
    message = Text(Point(500, 350), str)
    message.draw(gragh)
#查询结点数量
def getnum(tree):
    num=0
    if(tree!=None):
        num=num+1
        num+=getnum(tree.lchild)
        num+=getnum(tree.rchild)
    return num
def exp4(gragh,ptree):
    str="4.该二叉树结点数为"+repr(getnum(ptree))
    message = Text(Point(500, 400), str)
    message.draw(gragh)
#查询叶子结点数量
def getleaf(tree):
    num=0
    if(tree!=None):
        num+=getleaf(tree.lchild)
        num+=getleaf(tree.rchild)
        if(tree.lchild==None and tree.rchild==None):
            return 1
    return num
def exp5(gragh,ptree):
    str="5.该二叉树叶子结点数为"+repr(getleaf(ptree))
    message = Text(Point(500, 450), str)
    message.draw(gragh)
#查询两个度的结点数量
def getTwo(tree):
    num=0
    if(tree!=None):
        num+=getTwo(tree.lchild)
        num+=getTwo(tree.rchild)
        if(tree.lchild!=None and tree.rchild!=None):
            num+=1
    return num
def exp6(gragh,ptree):
    str="6.该二叉树有两个度的结点有："+repr(getTwo(ptree))
    message = Text(Point(500, 500), str)
    message.draw(gragh)
#查询父亲结点，兄弟结点，子节点
def findFather(tree,name):
    if(tree!=None):
        if(tree.lchild!=None and tree.lchild.elementType==name):
            return tree
        if(tree.rchild!=None and tree.rchild.elementType==name):
            return tree
        if findFather(tree.rchild,name)!=None:
            return  findFather(tree.rchild,name)
        if findFather(tree.lchild,name)!=None:
            return  findFather(tree.lchild,name)
    return None
def info(tree,name):
    father=findFather(tree,name)
    if(father.lchild!=None and father.lchild.elementType==name):
        brother=father.rchild
        son1=father.lchild.lchild
        son2=father.lchild.rchild
    else :
        brother=father.lchild
        son1=father.rchild.lchild
        son2=father.rchild.rchild
    return father,brother,son1,son2
def exp7(gragh ,tree,name):
    str=""
    father,brother,son1,son2=info(tree,name)
    if(father==None):
        str+="8.父节点不存在"+"\n"
    else:
        str+="8.父节点为："+repr(father.elementType)+"\n"
    if(brother==None):
        str+="兄弟结点不存在"+"\n"
    else:
        str+="兄弟结点为："+repr(brother.elementType)+"\n"
    if(son1==None):
        str+="左子结点不存在"+"\n"
    else:
        str+="左子结点为："+repr(son1.elementType)+"\n"
    if (son1 == None):
        str += "右子结点不存在"+"\n"
    else:
        str+="右子结点为："+repr(son2.elementType)+"\n"
    message = Text(Point(400, 550), str)
    message.draw(gragh)
#查询指定结点深度
def getdeep(tree,name,deep):   #EXP8
    if (tree != None):
        if (tree.elementType==name):
            return  deep
        deepleft=getdeep(tree.lchild,name,deep+1)
        deepright=getdeep(tree.rchild,name,deep+1)
        if(deepleft!=0):
            return  deepleft
        if(deepright!=0):
            return  deepright
    return 0
def exp8(gragh,tree,name):
    deep=getdeep(tree,name,1)
    if(deep==0):
        message = Text(Point(400, 500), "7.本结点不存在")
        message.draw(gragh)
        return 0
    else:
        message = Text(Point(400, 500), "7.本结点："+repr(name)+"深度为"+repr(deep))
        message.draw(gragh)
        return 1
#顺序存储变为二叉链表存储
def seqToNode(s,tree,i):
    if(tree!=None):
        if(i*2>len(s)):
            tree.lchild=None
        elif (s[i*2]==None) :
            tree.lchild=None
        else:
            temp = BiTree(s[i*2])
            tree.lchild=temp
        if(i*2+1>len(s)):
            tree.rchild=None
        elif (s[i*2+1]==None):
            tree.rchild=None
        else:
            temp = BiTree(s[i*2 + 1])
            tree.rchild=temp
        seqToNode(s,tree.lchild,i*2)
        seqToNode(s,tree.rchild,i*2+1)
#交换左右二叉树
def change(tree):#     EXP10
    if(tree!=None):
        temp=tree.lchild
        tree.lchild=tree.rchild
        tree.rchild=temp
    if(tree.lchild!=None):
        change(tree.lchild)
    if(tree.rchild!=None):
        change(tree.rchild)
#找所有结点的路径
def road(s,tree,all):
    if(tree.rchild==None and tree.lchild==None):
        all.append(repr(tree.elementType))
        all.append("到根节点的路径为"+reverse1(s)+"A"+"\n")
    if(tree.lchild!=None):
        road(s+tree.lchild.elementType,tree.lchild,all)
    if(tree.rchild!=None):
        road(s+(tree.rchild.elementType),tree.rchild,all)
def exp12(all,gragh,ptree):
    str=""
    road(str, ptree, all)
    for i in range (len(all)):
        str+=all[i]
    message = Text(Point(150, 450), "12.从每个叶子结点到根结点的路径:\n"+str)
    message.draw(gragh)
#按层次打印结点
def exp13(gragh,tree):
    que=Queue()
    que.enqueue(tree)
    str="13.结点按层次打印结果为\n"
    while(not que.isEmpty()):
        if(que.getTop().lchild!=None):
            que.enqueue(que.getTop().lchild)
        if(que.getTop().rchild!=None):
            que.enqueue(que.getTop().rchild)
        str+=repr(que.getTop().elementType)
        que.outqueue()
    message = Text(Point(600, 400), str)
    message.draw(gragh)
#查找最长路径
def maxpath(temp,tree,path,deep):
    if (tree.rchild == None and tree.lchild == None):
        return temp,deep
    deep1,deep2=0,0
    if (tree.lchild != None):
        path1,deep1=maxpath(temp + tree.lchild.elementType, tree.lchild, path,deep+1)
    if (tree.rchild != None):
        path2,deep2=maxpath(temp + (tree.rchild.elementType), tree.rchild, path,deep+1)
    if(deep1>deep2):
        return  path1,deep1
    else :
        return  path2,deep2
def exp15(gragh,tree):
    temp=""
    path=""
    path,deep=maxpath(temp,tree,path,1)
    message = Text(Point(600, 500), "15.该二叉树最长路径为：\nA"+path+"\n长度为"+repr(deep))
    message.draw(gragh)

值得注意的是13题需要用到队列，需要提前写好队列的代码如下：

class Queue:
    """模拟队列"""
    def __init__(self):
        self.items = []
    def isEmpty(self):
        return self.items == []
    def enqueue(self, item):
        self.items.insert(0, item)
    def outqueue(self):
        return self.items.pop()
    def size(self):
        return len(self.items)
    def getTop(self):
        return self.items[self.size()-1]

上述便是代码的全部代码，效果如下