03-樹(shù)3 Tree Traversals Again (25分)
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
[圖片上傳失敗...(image-1897-1511628506569)]Figure 1
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N
N
N (≤
3
0
\le 30
≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N
N
N). Then 2
N
2N
2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.
Output Specification:
For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MaxSize 30
struct{
int top;
int data[MaxSize];
}stack;
int preorder[MaxSize],inorder[MaxSize];
void GetPostOrder(int preorder[],int a,int b,int inorder[],int c,int d)
{
int i;
for( i=c;i<=d&&preorder[a]!=inorder[i];i++);
if(i>0) GetPostOrder(preorder,a+1,i,inorder,c,i-1);
if(i<d) GetPostOrder(preorder,i+1,b,inorder,i+1,d);
printf("%d ",inorder[i]);
}
int main()
{
freopen("in.txt","r",stdin);
stack.top=-1;
int n,pretop=0,intop=0,temp;
scanf("%d",&n);
for(int i=0;i<2*n;i++)
{
char s[5];
scanf("%s",s);
if(strcmp(s,"Push")==0){
scanf("%d",&temp);
stack.data[++stack.top]=temp;
preorder[pretop++]=temp;
}
else{
inorder[intop++]=stack.data[stack.top--];
}
}
GetPostOrder(preorder,0,n-1,inorder,0,n-1);
}