用python实现二叉树的基本操作

二叉树基本操作:

  • 创建二叉树 (need)
  • 递归遍历二叉树 (need)
    1)前序遍历:“根-左-右”
    2)中序遍历:“左-根-右”
    3)后序遍历:“左-右-根”
  • 非递归遍历二叉树
    1)前序遍历
    2)中序遍历
    3)后序遍历

创建二叉树,如下图:用python实现二叉树的基本操作_第1张图片

手动创建二叉树:
	方法一:
    A, B, C, D, E, F = [BinaryTreeNode(x) for x in '123456']
    A.left, A.right = B, C
    B.left, B.right = D, E
    C.right = F
    方法二:
    D = BinaryTreeNode(4)
    E = BinaryTreeNode(5)
    F = BinaryTreeNode(6)
    C = BinaryTreeNode(3, right=F)
    B = BinaryTreeNode(2, left=D, right=E)
    root = BinaryTreeNode(1, left=B, right=C)
    二叉树结点:
    class BinaryTreeNode:
	    def __init__(self, data=None, left=None, right=None):
	        self.data = data
	        self.left = left
	        self.right = right

遍历结果如下:
1)前序遍历:1、2、4、5、3、6
2)中序遍历:4、2、5、1、3、6
3)后序遍历:4、5、2、6、3、1


完整代码,如下:

# !usr/bin/env python  
# -*- coding:utf-8 -*-
"""
二叉树的基本操作
@author: YAOTIANLONG 
@file: DS_tree.py 
@time: 2019/10/28 
"""
class BinaryTreeNode:
    """二叉树结点"""
    def __init__(self, data=None, left=None, right=None):
        self.data = data
        self.left = left
        self.right = right


class BinaryTree:
    """
    二叉树基本操作:
        - 创建二叉树
        - 判断二叉树是否为空
        - 二叉树递归遍历:前序、中序、后序
        - 二叉树非递归遍历:前序、中序、后序
    """
    def __init__(self, root=None):
        self.root = root

    def is_empty(self):
        return True if self.root == None else False

    def preOrder(self, BinaryTreeNode):
        """递归
        前序遍历
        :param BinaryTreeNode:
        :return:
        """
        if BinaryTreeNode == None:
            return
        print(BinaryTreeNode.data, end=" ")
        self.preOrder(BinaryTreeNode.left)
        self.preOrder(BinaryTreeNode.right)

    def inOrder(self,BinaryTreeNode):
        """递归
        中序遍历
        :param BinaryTreeNode:
        :return:
        """
        if not BinaryTreeNode:
            return
        self.inOrder(BinaryTreeNode.left)
        print(BinaryTreeNode.data, end=" ")
        self.inOrder(BinaryTreeNode.right)

    def postOrder(self,BinaryTreeNode):
        """递归
        后序遍历
        :param BinaryTreeNode:
        :return:
        """
        if not BinaryTreeNode:
            return
        self.postOrder(BinaryTreeNode.left)
        self.postOrder(BinaryTreeNode.right)
        print(BinaryTreeNode.data, end=" ")

    # 非递归遍历,参考博客https://blog.csdn.net/z_ryan/article/details/80854233
    def preOrder1(self):
        """
        非递归前序遍历:
        当p非空或stack非空时,输出p.data并将p压入stack,遍历左子树(边遍历边打印,并把根节点存入栈中)
        当无左子树时,弹出stack,遍历该节点的右子树
        :param self:
        :return:
        """
        if self.root == None:
             return
        stack = []
        p = self.root
        while p != None or stack:
            while p != None:  # 边遍历边打印,并存入栈中,以后需要借助这些根节点进入右子树
                print(p.data, end=' ')
                stack.append(p)
                p = p.left
            if stack:  # 当p为空时,说明根和左子树都遍历完了,该进入右子树了
                p = stack.pop().right

    def inOrder1(self):
        """
        非递归中序遍历:
        先将p和左子树压入stack,当p为空时,弹出stack并输出数据,转而将p变为右子树
        :return:
        """
        if self.root == None:  # 空树
            return
        stack = []
        p = self.root
        while p != None or stack:
            while p != None:  # 一直遍历到左子树最下边,边遍历边保存根节点到栈中
                stack.append(p)
                p = p.left
            if stack:  # 当p为空时,说明已经到达左子树最下边,这时需要出栈了
                s = stack.pop()
                print(s.data, end=' ')
                p = s.right  # 进入右子树,开始新的一轮左子树遍历(这是递归的自我实现)


    def postOrder1(self):
        """
        非递归后序遍历
        只有当左子树和右子树都被访问了,才能访问根节点
        当当前根节点右子树和per相等时,说明已经可以访问该根节点了
        # https://blog.csdn.net/qq_33951180/article/details/52687692
        :return:
        """
        if self.root == None:
            return
        stack = []
        p,per = self.root,None
        while p != None or stack:
            while p != None:
                stack.append(p)
                p = p.left
            top = stack[-1]  # 栈顶元素,p指针回退
            if top.right == None or top.right == per:
                # top存的是当前的根节点,当top的右子树为空或者top的右子树为pos,说明右子树已经遍历过,
                # 这时就可以访问当前的根节点了
                print(top.data,end=' ')
                per = top
                stack.pop()
            else:
                p = top.right


if __name__ == "__main__":
    #先创建叶节点
    # root, B, C, D, E, F = [BinaryTreeNode(x) for x in '123456']
    # root.left, root.right = B, C
    # B.left, B.right = D, E
    # C.right = F

    D = BinaryTreeNode(4)
    E = BinaryTreeNode(5)
    F = BinaryTreeNode(6)
    C = BinaryTreeNode(3, right=F)
    B = BinaryTreeNode(2, left=D, right=E)
    root = BinaryTreeNode(1, left=B, right=C)
    # 遍历
    bt = BinaryTree(root)
    print('前序:')
    bt.preOrder(bt.root)
    print()
    bt.preOrder1()
    print()
    print('中序:')
    bt.inOrder(bt.root)
    print()
    bt.inOrder1()
    print()
    print('后序:')
    bt.postOrder(bt.root)
    print()
    bt.postOrder1()
输出结果:
C:\Software\Anaconda3\envs\tf_gpu\python.exe D:/PycharmProjects/剑指offer/DS_tree.py
前序:
1 2 4 5 3 6 
1 2 4 5 3 6 
中序:
4 2 5 1 3 6 
4 2 5 1 3 6 
后序:
4 5 2 6 3 1 
4 5 2 6 3 1 
Process finished with exit code 0

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