• 基础数据结构 之 树(python实现)


    树是数据结构中常用到的一种结构,其实现较栈和队稍为复杂一些。若树中的所有节点的孩子节点数量不超过2个,则该为一个二叉树。二叉树可用于查找和排序等。二叉树的主要操作有:建树,遍历等。遍历是树中的一个最为重要的操作,可分为深度优先遍历和广度优先遍历。其中,尝试优先遍历又可分为先序遍历,中序遍历和后序遍历。深度优先遍历可使用递规来实现,也可以用栈和队通过循环实现。后序的非递规遍历,比其他两种遍历稍为复杂些。

    下面给出一个python实现二叉树的例子:

    class Node(object):
        def __init__(self, data = -1, lchild = None, rchild = None):
            self.data = data
            self.lchild = lchild
            self.rchild = rchild
    
    class BinaryTree(object):
        def __init__(self):
            self.root = Node()
    
        def add(self, data):
            node = Node(data)
            if self.isEmpty():
                self.root = node
            else:
                tree_node = self.root
                queue = []
                queue.append(self.root)
    
                while queue:
                    tree_node = queue.pop(0)
                    if tree_node.lchild == None:
                        tree_node.lchild = node
                        return
                    elif tree_node.rchild == None:
                        tree_node.rchild = node
                        return
                    else:
                        queue.append(tree_node.lchild)
                        queue.append(tree_node.rchild)
    
        def pre_order(self, start):
            node = start
            if node == None:
                return
    
            print node.data,
            if node.lchild == None and node.rchild == None:
                return
            self.pre_order(node.lchild)
            self.pre_order(node.rchild)
    
        def pre_order_loop(self):
            if self.isEmpty():
                return
    
            stack = []
            node = self.root
            while node or stack:
                while node:
                    print node.data,
                    stack.append(node)
                    node = node.lchild
                if stack:
                    node = stack.pop()
                    node = node.rchild
    
        def in_order(self, start):
            node = start
            if node == None:
                return
            self.in_order(node.lchild)
            print node.data,
            self.in_order(node.rchild)
    
        def in_order_loop(self):
            if self.isEmpty():
                returen
            
            stack = []
            node = self.root
            while node or stack:
                while node:
                    stack.append(node)
                    node = node.lchild
    
                if stack:
                    node = stack.pop()
                    print node.data,
                    node = node.rchild
    
        def post_order(self, start):
            node = start
            if node == None:
                return
            self.post_order(node.lchild)
            self.post_order(node.rchild)
            print node.data,
    
        
        def post_order_loop(self):
            if self.isEmpty():
                return
            
            node = self.root
            stack = []
            queue = []
            queue.append(node)
            while queue:
                node = queue.pop()
                if node.lchild:
                    queue.append(node.lchild)
                if node.rchild:
                    queue.append(node.rchild)
                stack.append(node)
            while stack:
                print stack.pop().data,
    
        #if lchild and rchild are None or lchild and rchild are printed, print the parent node node and pop out of the stack
        #else lchild and rchild push into the stack
        def post_order_loop1(self):
            if self.isEmpty():
                return
    
            stack = []
            top = -1
            node = self.root
            stack.append(node)
            #we need to recognize the last printed node
            top += 1
            pre = None
            while stack:
                node = stack[-1]
                if node.lchild is None and node.rchild is None:
                    print node.data,
                    pre = node
                    top -= 1
                elif not pre and (node.lchild == pre or node.rchild == pre):
                    print node.data,
                    pre = node
                    top -= 1
                else:
                    if node.rchild:
                        if top < len(stack)-1:
                            stack[top] = node.rchild
                        else:
                            stack.append(node.rchild)
                    if node.lchild:
                        if top < len(stack)-1:
                            stack[top] = node.lchild
                        else:
                            stack.append(node.lchild)
    
        def level_order(self):
            node = self.root
            if node == None:
                return
            
            queue = []
            queue.append(node)
    
            while queue:
                node = queue.pop(0)
                print node.data,
                if node.rchild:
                    queue.append(node.rchild)
                if node.lchild:
                    queue.append(node.lchild)
            print
    
        def isEmpty(self):
            return True if self.root.data == -1 else False
    
    if __name__ == '__main__':
        arr = []
        for i in range(10):
            arr.append(i)
        print arr
    
        tree = BinaryTree()
        for i in arr:
            tree.add(i)
        print 'level_order:'
        tree.level_order()
        print 'pre order:'
        tree.pre_order(tree.root)
        print '
    pre order loop:'
        tree.pre_order_loop()
        print '
    in_order:'
        tree.in_order(tree.root)
        print '
    in_order loop:'
        tree.in_order_loop()
        print '
    post_order:'
        tree.post_order(tree.root)
        print '
    post_order_loop:'
        tree.post_order_loop()
        print '
    post_order_loop1:'
        tree.post_order_loop1()
  • 相关阅读:
    Ant中如何添加第三方jar包依赖
    蓝绿部署、滚动部署、灰度发布(金丝雀发布)
    通过 Docker 部署 nodejs egg 项目
    MySql 语句执行顺序
    MySql 语句执行顺序
    Laravel验证字段是否由中文、英文组成
    在 KubeSphere 中部署高可用 Redis 集群
    一文读懂 Prometheus 长期存储主流方案
    基于 CoreDNS 和 K8s 构建云原生场景下的企业级 DNS
    NebulaGraph 的云产品交付实践
  • 原文地址:https://www.cnblogs.com/sunada2005/p/3326160.html
Copyright © 2020-2023  润新知