1138 Postorder Traversal （25 分）（根据先序和中序序列，后序遍历二叉树）

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本文介绍了一个算法问题，即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。通过递归地构建二叉树并输出根节点的方式，实现了这一目标。

1138 Postorder Traversal （25 分）

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:

#include <bits/stdc++.h>

using namespace std;
vector<int> pre,in;
bool flag=false;
void postOrder(int prel,int inl,int inr)
{
    if(inl>inr||flag==true)
        return;
    int i=inl;
    while(in[i]!=pre[prel]) i++;
    postOrder(prel+1,inl,i-1);
    postOrder(prel+i-inl+1,i+1,inr);
    if(flag==false)
    {
        printf("%d",in[i]);
        flag=true;
    }
}
int main()
{
    int n;
    scanf("%d",&n);
    pre.resize(n);
    in.resize(n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&pre[i]);
    }
    for(int i=0;i<n;i++)
    {
        scanf("%d",&in[i]);
    }
    postOrder(0,0,n-1);
    return 0;
}