HDU1710 Binary Tree Traversals如何通过遍历二叉树实现？

本文共计702个文字，预计阅读时间需要3分钟。

Binary Tree TraversalsTime Limit: 1000MSMemory Limit: 32768KB

64bit IO Format: %I64d %I64u

Submit Status DescriptionA binary tree is a finite set of vertices that is either empty or consists of a root node r and two disjoint binary trees called the left subtree and the right subtree.

Binary Tree Traversals

Time Limit:1000MSMemory Limit:32768KB64bit IO Format:%I64d & %I64u

Submit Status

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

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

Sample Output

7 4 2 8 9 5 6 3 1

#include<stdio.h> #include<string.h> #include<stdlib.h> typedef struct tnode { int data; tnode *left, *right; } TNODE2; TNODE2 *make(int *pstr,int *istr,int nn) { TNODE2 *ptr; int *rstr; int k; if(nn<=0) return NULL; ptr = (TNODE2*)malloc(sizeof(TNODE2)); ptr->data = *pstr; for(rstr=istr; rstr<istr+nn; rstr++) if(*rstr == *pstr) break; k = rstr - istr; ptr->left = make(pstr+1,istr,k); ptr->right = make(pstr+k+1,istr+k+1,nn-k-1); return ptr; } int h = 0; void lastorder(TNODE2 *t,int n) { if(t == NULL) return ; lastorder(t->left,n); lastorder(t->right,n); h++; if(h == n) { printf("%d\n",t->data); } else { printf("%d ",t->data); } } int main() { TNODE2 *tree; int a[10001],b[10010]; int n; while(scanf("%d",&n)!=EOF) { h = 0; for(int i=0;i<n;i++) { scanf("%d",&a[i]); } for(int i=0;i<n;i++) { scanf("%d",&b[i]); } tree = make(a,b,n); lastorder(tree,n); } return 0; }