二叉树前序中序遍历得到后序遍历

Posted 2020-10-30 每天学习一点点

二叉树的前序遍历为:{1,2,4,7,3,5,6,8}，中序遍历为:{4,7,2,1,5,3,8,6}，求后序遍历

　

 

 

# -*- coding:utf-8 -*-
class Node:
    def __init__(self, x):
        self.data = x
        self.lchild = None
        self.rchild = None
class Solution:

    def __init__(self):
        self.list=[]

    def reConstructBinaryTree(self, pre, tin):
        if len(tin)==0 or len(pre)==0:
            self.list.append(Node(-1))
            return Node(-1)
        data = pre.pop(0)
        root=Node(data)
        index=tin.index(data)
        #递归得到后序遍历
        root.lchild=self.reConstructBinaryTree(pre,tin[0:index])
        root.rchild=self.reConstructBinaryTree(pre,tin[index+1:])
        self.list.append(root)
        return root

    #层次遍历
    def level_queueAndStack(self,root):
        if root==None:
            return
        stack_1=[]
        stack_2=[]
        stack_1.append(root)
        stack_2.append(root)
        while stack_1:
            node=stack_1.pop(0)
            if node.lchild:
                stack_1.append(node.lchild)
                stack_2.append(node.lchild)
            if node.rchild:
                stack_1.append(node.rchild)
                stack_2.append(node.rchild)
        while stack_2:
            print stack_2.pop(0).data,

s=Solution()
s.reConstructBinaryTree([1,2,4,7,3,5,6,8],[4,7,2,1,5,3,8,6])

#输出树的后序遍历
l=[i.data for i in s.list if i.data!=-1]
print l

s.level_queueAndStack(s.list.pop())

 输出结果：

[7, 4, 2, 5, 8, 6, 3, 1]
1 2 3 4 -1 5 6 -1 7 -1 -1 8 -1 -1 -1 -1 -1