主要是:前序遍历、中序遍历、后序遍历、层级遍历、非递归前序遍历、非递归中序遍历、非递归后序遍历

复制代码 代码如下:
#!/usr/bin/env python
#-*- coding:utf8 -*-


class TreeNode(object):
    def __init__(self, data=None, left=None, right=None):
        self.data = data
        self.left = left
        self.right = right


class Tree(object):
    def __init__(self, root=None):
        self.root = None

    def makeTree(self, data, left, right):
        self.root = TreeNode(data, left, right)

    def is_empty(self):
        """是否为空 """
        if self.root is None:
            return True
        return False

    def preOrder(self, r):
        """前序遍历 """
        if not r.is_empty():
            print r.root.data
            if r.root.left is not None:
                r.preOrder(r.root.left)
            if r.root.right is not None:
                r.preOrder(r.root.right)

    def inOrder(self, r):
        """中序遍历 """
        if not r.is_empty():
            if r.root.left is not None:
                r.preOrder(r.root.left)
            print r.root.data
            if r.root.right is not None:
                r.preOrder(r.root.right)

    def postOrder(self, r):
        """后续遍历 """
        if not r.is_empty():
            if r.root.left is not None:
                r.preOrder(r.root.left)
            print r.root.data
            if r.root.right is not None:
                r.preOrder(r.root.right)

    def levelOrder(self, r):
        """层级遍历 """
        if not r.is_empty():
            s = [r]
            while len(s) > 0:
                temp = s.pop(0)  # 先弹出最先append到的点
                if temp and temp.root is not None:
                    print temp.root.data
                    if temp.root.left is not None:
                        s.append(temp.root.left)
                    if self.root.right is not None:
                        s.append(temp.root.right)

    def preOrder1(self, r):
        """非递归 前序遍历 """
        stack = []
        current = r
        while len(stack) > 0 or (current and not current.is_empty()):
            while current and not current.is_empty():
                print current.root.data
                stack.append(current)
                current = current.root.left
            if len(stack) > 0:
                current = stack.pop()
                current = current.root.right

    def inOrder1(self, r):
        """非递归 中序遍历 """
        stack = []
        current = r
        while len(stack) > 0 or (current and not current.is_empty()):
            while current and not current.is_empty():
                stack.append(current)
                current = current.root.left
            if len(stack) > 0:
                current = stack.pop()
                print current.root.data
                current = current.root.right

    def postOrder1(self, r):
        """非递归 后续遍历 """
        stack = []
        current = r
        pre = None
        while len(stack) > 0 or (current and not current.is_empty()):
            if current and not current.is_empty():
                stack.append(current)
                current = current.root.left
            elif stack[-1].root.right != pre:
                current = stack[-1].root.right
                pre = None
            else:
                pre = stack.pop()
                print pre.root.data

    def leaves_count(self, r):
        """求叶子节点个数 """
        if r.is_empty():
            return 0
        elif (not r.root.left) and (not r.root.right):
            return 1
        else:
            return r.root.left.leaves_count(r.root.left) + r.root.right.leaves_count(r.root.right)


if __name__ == '__main__':
    """二叉树"""
    ra, rb, rc, rd, re, rf = Tree(), Tree(), Tree(), Tree(), Tree(), Tree()
    ra.makeTree("a", None, None)
    rb.makeTree("b", None, None)
    rc.makeTree("c", None, None)
    rd.makeTree("d", None, None)
    re.makeTree("e", None, None)
    rf.makeTree("f", None, None)
    r1, r2, r3, r4, r = Tree(), Tree(), Tree(), Tree(), Tree()
    r1.makeTree("-", rc, rd)
    r2.makeTree("*", rb, r1)
    r3.makeTree("+", ra, r2)
    r4.makeTree("/", re, rf)
    r.makeTree("-", r3, r4)
    r.preOrder(r)
    r.inOrder(r)
    r.postOrder(r)
    r.levelOrder(r)
    print r.leaves_count(r)


大学的时候学过kmp算法,最近在看的时候发现竟然忘了,所以去重新看了看书,然后用python写下了这个算法:
复制代码 代码如下:
def kmp(text, pattern):
    """kmp算法 """
    pattern = list(pattern)
    next = [-1] * len(pattern)
    #next 函数
    i, j = 1, -1
    for i in range(1, len(pattern)):
        j = next[i - 1]
        while True:
            if pattern[i - 1] == pattern[j] or j == -1:
                next[i] = j + 1
                break
            else:
                j = next[j]
    #循环比较
    i, j = 0, 0
    while i < len(text) and j < len(pattern):
        if text[i] == pattern[j] or j == -1:
            i += 1
            j += 1
        else:
            j = next[j]
    #返回结果 如果匹配,返回匹配的位置,否则返回-1
    if j == len(pattern):
        print i – j
    else:
        print -1

广告合作:本站广告合作请联系QQ:858582 申请时备注:广告合作(否则不回)
免责声明:本站资源来自互联网收集,仅供用于学习和交流,请遵循相关法律法规,本站一切资源不代表本站立场,如有侵权、后门、不妥请联系本站删除!

《魔兽世界》大逃杀!60人新游玩模式《强袭风暴》3月21日上线

暴雪近日发布了《魔兽世界》10.2.6 更新内容,新游玩模式《强袭风暴》即将于3月21 日在亚服上线,届时玩家将前往阿拉希高地展开一场 60 人大逃杀对战。

艾泽拉斯的冒险者已经征服了艾泽拉斯的大地及遥远的彼岸。他们在对抗世界上最致命的敌人时展现出过人的手腕,并且成功阻止终结宇宙等级的威胁。当他们在为即将于《魔兽世界》资料片《地心之战》中来袭的萨拉塔斯势力做战斗准备时,他们还需要在熟悉的阿拉希高地面对一个全新的敌人──那就是彼此。在《巨龙崛起》10.2.6 更新的《强袭风暴》中,玩家将会进入一个全新的海盗主题大逃杀式限时活动,其中包含极高的风险和史诗级的奖励。

《强袭风暴》不是普通的战场,作为一个独立于主游戏之外的活动,玩家可以用大逃杀的风格来体验《魔兽世界》,不分职业、不分装备(除了你在赛局中捡到的),光是技巧和战略的强弱之分就能决定出谁才是能坚持到最后的赢家。本次活动将会开放单人和双人模式,玩家在加入海盗主题的预赛大厅区域前,可以从强袭风暴角色画面新增好友。游玩游戏将可以累计名望轨迹,《巨龙崛起》和《魔兽世界:巫妖王之怒 经典版》的玩家都可以获得奖励。