杭电1710 Binary Tree Traversals（二叉树的遍历）

Binary Tree Traversals

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3684    Accepted Submission(s): 1628

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 

/*加油！！！Time:2014-11-29 13:55*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAX=2005;typedef struct Tree{int num;Tree *lChild;Tree *rChild;}*tree;int pre[MAX],in[MAX],post[MAX];void Creat(tree &rt,int *pre,int *in,int len){//if(len==0)rt==NULL; //else{int p=-1;for(int i=0;i<len;i++){if(in[i]==pre[0]){p=i;break;}}//if(p==-1) rt=NULL;//else{rt=new Tree;rt->num=pre[0];if(p==0){rt->lChild=NULL;}else{Creat(rt->lChild,pre+1,in,p);//传入的是左右子树的第一个节点 //传入根节点，前序的根节点，中序的根节点，树的长度 }if(p==len-1){rt->rChild=NULL;}else{Creat(rt->rChild,pre+p+1,in+p+1,len-p-1);//右子树的第一个位置为pos+1 }//}//}}void Print(tree &rt,int &k){if(rt==NULL) return;Print(rt->lChild,k);Print(rt->rChild,k);post[k++]=rt->num;free(rt);}int main(){int n,i,j,k;while(scanf("%d",&n)!=EOF){memset(pre,0,sizeof(pre));memset(in,0,sizeof(in));memset(post,0,sizeof(post));for(i=0;i<n;i++)scanf("%d",&pre[i]);for(i=0;i<n;i++)scanf("%d",&in[i]);tree rt;k=0;Creat(rt,pre,in,n);Print(rt,k);//printf("k=%d\n",k);printf("%d",post[0]);for(i=1;i<k;i++)printf(" %d",post[i]);puts("");}return 0;}