PAT (Advanced Level) Practice 1138 Postorder Traversal （25 分）树的重构

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最新推荐文章于 2022-02-04 16:29:41 发布

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本文介绍了一种算法，通过给定的前序和中序遍历序列，求解二叉树后序遍历的第一个节点值。具体实现包括读取输入数据，构建前序和中序遍历序列，并使用递归方法找到后序遍历的首个元素。

Suppose that all the keys in a binary tree are distinct positive integers. Given the preorder and inorder traversal sequences, you are supposed to output the first number of the postorder 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 (≤ 50,000), the total number of nodes in the binary tree. The second line gives the preorder 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 first number of the postorder traversal sequence of the corresponding binary tree.

Sample Input:

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

Sample Output:

题意

给出树的前序和中序，求后序的第一个节点的值。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;
const int maxn=5*1e4+5;
int n;
int flag=0;
vector<int>pre;
vector<int>in;
vector<int>post;
map<int,int>ma;
int pos=0;
void Post (int l,int r)
{
    if(post.size())
    {
        flag=1;
        return;
    }
    if(flag)
    {
        return;
    }
    if(l>r)
    {
        return;
    }
    int temp=pre[pos++];
    int loc=ma[temp];
    Post(l,loc-1);
    Post(loc+1,r);
    post.push_back(temp);
}
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
    {
        int x;
        scanf("%d",&x);
        pre.push_back(x);
    }
    for (int i=0;i<n;i++)
    {
        int x;
        scanf("%d",&x);
        in.push_back(x);
        ma[x]=i;
    }
    Post(0,n-1);
    printf("%d\n",post[0]);
    return 0;
}