Ideas

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Search in the middle-order sequence through the sequence of the pre-order sequence, and use the characteristics of the middle-order sequence to divide and conquer to both sides, first divide and conquer the left and then divide and conquer the right, and finally output the root node;
this idea will have many repeated solutions, So we need to borrow the visited array to remove unnecessary duplicate solutions, and then we can get the output of the post-order sequence without building a binary tree.

Algorithm implementation

#include<iostream>
using namespace std;
#define N 100
//    1
//  2   3
//#  4 5 6
// 前序 1 2 4 3 5 6
// 中序 2 4 1 5 3 6
// 后序 4 2 5 6 3 1
/**
    辅助方法
    在序列中查找
    @param  int[]   序列
    @param  int     查询开始位置
    @param  int     查询结束位置
    @param  int     待查询元素
    @return int     查找到的元素的下标
*/
int search(int[],int,int,int);
/**
    根据前序和中序序列输出后序序列的递归方法
    @param  int     前序序列的长度
    @param  int[]   前序序列
    @param  int     遍历前序序列的开始位置
    @param  int[]   中序序列
    @param  int     中序序列递归开始位置
    @param  int     中序序列递归结束位置
*/
void postOrderSeq(int,int[],int,int[],int,int);

int visited[N]={
    
    0}; //防止重复访问

int main(){
    
    
    int preOrderSeq[N] = {
    
    1,2,4,3,5,6},inOrderSeq[N] = {
    
    2,4,1,5,3,6};
    int len = 6;
    postOrderSeq(len,preOrderSeq,0,inOrderSeq,0,len);
}

int search(int inOrderSeq[],int inStart,int inEnd,int x){
    
    
    for(int i = inStart;i < inEnd;i++)
        if(inOrderSeq[i] == x)return i;
    return -1;
}

void postOrderSeq(int len,int preOrderSeq[],int preStart,int inOrderSeq[],int inStart,int inEnd){
    
    
    for(int i = preStart;i < len;i++){
    
    
        int index = search(inOrderSeq,inStart,inEnd,preOrderSeq[i]);
        if(index != -1&&visited[index]!=1) {
    
    
            visited[index] = 1;
            postOrderSeq(len,preOrderSeq,i,inOrderSeq,0,index);
            postOrderSeq(len,preOrderSeq,i,inOrderSeq,index+1,inEnd);
            cout<<preOrderSeq[i]<<" ";
        }
    }
}

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(Recursive method) Do not build a binary tree model to solve the problem of "write out the post-order sequence according to the pre-order sequence and middle-order sequence"

(Recursive method) Do not build a binary tree model to solve the problem of "write out the post-order sequence according to the pre-order sequence and middle-order sequence"

Ideas

Algorithm implementation

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