HDU 1710 已知前序中序须求后序。

Binary Tree Traversals

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5335    Accepted Submission(s): 2470

Problem Description

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.

In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.

In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.

In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.

Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.

 

Input

The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.

 

Output

For each test case print a single line specifying the corresponding postorder sequence.

 

Sample Input

91 2 4 7 3 5 8 9 64 7 2 1 8 5 9 3 6

 

Sample Output

7 4 2 8 9 5 6 3 1

 

Source

HDU 2007-Spring Programming Contest

 

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后序是左右根。

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>using namespace std;int a[1005];int b[1005];void printf(int begin,int end,int mid){    int i;    if(begin>end)        return;    for(i=begin; i<=end; i++)        if(b[i]==a[mid])            break;    printf(begin,i-1,mid+1);  //中序，先左，后跟，再右。此时i为跟节点，左子树必定在begin到i-1里面。  mid+1是因为前序的左子树为mid+1,此时作为根，    printf(i+1,end,mid-begin+i+1);//中序，先左，后跟，再右。此时i为跟节点，右子树必定在i+1到end里面。 同理    cout<<a[mid];    if(mid==1)        cout<<endl;    else        cout<<" ";}int main(){    int n;    while(cin>>n)    {        int i;        for(i=1; i<=n; i++)            cin>>a[i];        for(i=1; i<=n; i++)            cin>>b[i];        printf(1,n,1);    }    return 0;}