hdu1710 Binary Tree Traversals

Binary Tree Traversals

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4995    Accepted Submission(s): 2278

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

通过先序遍历和中序遍历输出后续遍历。

因为先序遍历会优先输出根节点，所以可以利用这个性质将在中序遍历中划分出左右子树，依照这个性质递归输出节点就好了

#include <iostream>#include <cstdio>using namespace std;int a[1005], b[1005], n;void func(int beg, int end, int node)   //beg, end表示当前树在数组b上的区间, node表示当前节点在a中的下标{    if (beg < 0 || end >n || beg > end)        return;    for (int i = beg; i <= end; ++i)    {        if (a[node] == b[i])        {            func(beg, i - 1, node + 1);            func(i + 1, end, node + i - beg + 1);            cout << a[node] << " ";            break;        }    }}int main(){    while (cin >> n)    {        int node;        for (int i = 1; i <= n; ++i)            cin >> a[i];        for (int i = 1; i <= n; ++i)        {            cin >> b[i];            if (a[1] == b[i])                node = i;        }        func(1, node - 1, 2);        func(node + 1, n, node + 1);        cout << a[1] << endl;    }}