HDU 1710 Binary Tree Traversals

Problem DescriptionA 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 traversedor 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. InputThe 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 assumethey are always correspond to a exclusive binary tree. OutputFor 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题目大意：给出二叉树的前序，中序，求出二叉树的后序首先要明白是什么二叉树，以及二叉树的前序，中序，后序分别是什么。这里有两个写的比较详细的博客，可以作为了解：关于二叉树：http://blog.csdn.net/luckyxiaoqiang/article/details/7518888关于前序中序得到后序http://www.cnblogs.com/rain-lei/p/3576796.html思路很容易想到，利用递归的思想，当遍历完左子树和右子树之后才输出根节点

代码如下：

#include<stdio.h>int pre[1001],in[1001];int flag;//在给定的区间内寻找子树的根节点int search(int value,int head,int tail){int i;for(i=head;i<=tail;i++){if(in[i]==value){return i;}}return -1; //在指定区间没找到返回－1}int branch(int head,int tail){int loca,value;loca=search(pre[flag],head,tail);value=pre[flag];if(loca==-1) //若返回－1说明子树为空return 0;flag++;branch(head,loca-1);//先遍历左子树branch(loca+1,tail);//再遍历右子树//最后输出根节点if(value!=pre[0])//后序遍历，根节点一定是在最后一个printf("%d ",value);elseprintf("%d",value);return 0;}int main(){int n,i;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)scanf("%d",&pre[i]);for(i=0;i<n;i++)scanf("%d",&in[i]);flag=0;branch(0,n-1);printf("\n");}}