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
第一个二叉树题
注意建法，proot=(Node)calloc()才可以修改，而这里不需要哨兵节点
注意递归，先终止项，再继续项
#include<stdio.h>
#include<stdlib.h>
#define N 1005
int pre[N],in[N];
int n,i;
typedef struct node{
int value;
struct node*pl,*pr;
}Node;
void build(int a,int b,int len,Node*proot){
//处理该点
proot->value=pre[a];//注意不是scanf
if(len==1){
return;
}
int cnt;
for(cnt=0;pre[a]!=in[b+cnt];cnt++);
if(cnt){
proot->pl=(Node*)calloc(1,sizeof(Node));
build(a+1,b,cnt,proot->pl);
}
if(len-cnt-1){
proot->pr=(Node*)calloc(1,sizeof(Node));
build(a+cnt+1,b+cnt+1,len-cnt-1,proot->pr);
}
}
void post(Node*proot,int flag){
if(!proot->pl&&!proot->pr){
printf("%d",proot->value);
if(flag)printf(" ");
return;
}
if(proot->pl){
post(proot->pl,1);
}
if(proot->pr){
post(proot->pr,1);
}
printf("%d",proot->value);
if(flag)printf(" ");
}
int main()
{
while(~scanf("%d",&n)){
Node*proot=(Node*)calloc(1,sizeof(Node));//注意要直接有一个头
for(i=1;i<=n;i++){
scanf("%d",&pre[i]);
}
for(i=1;i<=n;i++){
scanf("%d",&in[i]);
}
build(1,1,n,proot);
post(proot,0);
printf("\n");
}
return 0;
}