1020 Tree Traversals (25 分) 给出后序和中序遍历，输出层序遍历

本是古典何须时尚 2021-10-30 00:44 147阅读 0赞

1020 Tree Traversals (25 分)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

### Input Specification: ###

Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

### Output Specification: ###

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

### Sample Input: ###

7
    2 3 1 5 7 6 4
    1 2 3 4 5 6 7

### Sample Output: ###

4 1 6 3 5 7 2

突然发现这道题以前写过：[https://blog.csdn.net/TheWise\_lzy/article/details/79394852][https_blog.csdn.net_TheWise_lzy_article_details_79394852]

但是方法不一样 嘤嘤嘤

#include <iostream>
    #include <algorithm>
    #include <vector>
    using namespace std;
    struct node 
    {
        int index,value;
    }p;
    int post[100],in[100];
    vector<node> ans;
    bool cmp(node a, node b) 
    {
        return a.index<b.index;
    }
    void pre(int root,int start,int end,int index) 
    {
        if (start>end) return;
        int i=start;
        while(i<end&&in[i]!=post[root]) i++;
        p.index=index;
        p.value=post[root];
        ans.push_back(p);
        pre(root-1-end+i,start,i-1,2*index+1);
        pre(      root-1,  i+1,end,2*index+2);
    }
    int main() 
    {
        int n,i;
        scanf("%d", &n);
        for(i=0;i<n;i++) 
        scanf("%d",&post[i]);
        for(i=0;i<n;i++) 
        scanf("%d",&in[i]);
        pre(n-1,0,n-1,0);
        sort(ans.begin(),ans.end(),cmp);
        for(i=0;i<ans.size();i++) 
        {
            if(i)printf(" ");
            printf("%d",ans[i].value);
        }
        printf("\n");
    }

[https_blog.csdn.net_TheWise_lzy_article_details_79394852]: https://blog.csdn.net/TheWise_lzy/article/details/79394852