Binary Tree Traversals
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3475 Accepted Submission(s): 1555
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 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.
1 2 4 7 3 5 8 9 6
4 7 2 1 8 5 9 3 6
/*hdu 1710 二叉树*/
//#define LOCAL
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=;
int aa[maxn],bb[maxn];
void dfs(int a,int b,int n,int tag)
{
int i;
if(n<=)return ;
for(i=;aa[a]!=bb[b+i];i++);
dfs(a+,b,i,);
dfs(a+i+,b+i+,n-i-,);
printf("%d",aa[a]);
if(!tag)printf(" ");
}
int main()
{
#ifdef LOCAL
freopen("test.in","r",stdin);
#endif
int n,i,k,pre;
while(scanf("%d",&n)!=EOF)
{
for(i=;i<=n;i++)
scanf("%d",aa+i);
for(i=;i<=n;i++)
scanf("%d",bb+i);
dfs(,,n,);
printf("\n");
}
return ;
}
顺便做了一个二叉树知道其中两个序列求第三个序列的模板吧！ 注意： 知道前序和后序是无法求出唯一二叉树的！
代码：
//这部分检验aa[]为先序,bb[]为中序
void dfs_1(char aa[],char bb[], int a,int b,int n)
{
if(n<=)return ;
int i;
for(i=;aa[a]!=bb[b+i];i++);
dfs_1(aa,bb,a+,b,i); //左子树
dfs_1(aa,bb,a+i+,b+i+,n-i-); //右子树
printf("%c",aa[a]);
}
//aa[]为中序，bb[]为后序
void dfs_3(char aa[],char bb[],int a,int b,int n)
{
if(n<=)return ;
printf("%c",bb[b]);
int i;
for(i=;bb[b]!=aa[a+i];i++);
dfs_3(aa,bb,a,b-n+i,i); //左子树
dfs_3(aa,bb,a+i+,b-,n-i-); //右子树
}