1138. Postorder Traversal (25)【已知前序和中序求后序】

最新推荐文章于 2024-10-17 09:50:44 发布

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最新推荐文章于 2024-10-17 09:50:44 发布

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本文介绍了一个算法问题：给定二叉树的前序和中序遍历序列，求解对应的后序遍历序列的第一个数字。通过递归方法实现了这一目标，并提供了完整的C++代码实现。

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1138. Postorder Traversal (25)

时间限制

600 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 (<=50000), 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:

题解

#include <bits/stdc++.h>
#include<vector>
#define ms(x,y) memset(x,y,sizeof(x))
const int M=1e4+10;
using namespace std;
int j,k,n,m,q;
vector<int> pre, in;
bool flag=0;

void post(int prel,int inl,int inr)
{
    if(inl>inr||flag)return;
    int i=inl;
    while(in[i]!=pre[prel])i++;
    post(prel+1,inl,i-1);
    post(prel+i-inl+1,i+1,inr);
    if(!flag){
        printf("%d\n",in[i]);
        flag=1;
    }
}

int main()
{
    scanf("%d",&n);
    pre.resize(n);in.resize(n);
    for(int i=1;i<=n;i++)scanf("%d",&pre[i]);
    for(int i=1;i<=n;i++)scanf("%d",&in[i]);
    post(1,1,n);
    return 0;
}