浙大PAT：7-5 Tree Traversals Again (25分)

7-5 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.

Figure 1

Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 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.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

stack.h

#include <stdbool.h>
#include <malloc.h>
#define ERROR -1

typedef int ElementType;
typedef int Position;
struct SNode {
    ElementType *Data; /* 存儲元素的數組 */
    Position Top;      /* 棧頂指針 */
    int MaxSize;       /* 堆棧最大容量 */
};
typedef struct SNode* Stack;
 
Stack CreateStack( int MaxSize )
{
    Stack S = (Stack)malloc(sizeof(struct SNode));
    S->Data = (ElementType *)malloc(MaxSize * sizeof(ElementType));
    S->Top = -1;
    S->MaxSize = MaxSize;
    return S;
}
 
bool IsFull( Stack S )
{
    return (S->Top == S->MaxSize-1);
}
 
bool Push( Stack S, ElementType X )
{
    if ( IsFull(S) ) {
        return false;
    }
    else {
        S->Data[++(S->Top)] = X;
        return true;
    }
}
 
bool IsEmpty( Stack S )
{
    return (S->Top == -1);
}
 
ElementType Pop( Stack S )
{
    if ( IsEmpty(S) ) {
        return ERROR; /* ERROR是ElementType的特殊值，標誌錯誤 */
    }
    else 
        return ( S->Data[(S->Top)--] );
}

ElementType Top(Stack S){
	if(IsEmpty(S)){
		return ERROR;
	}
	else
		return S->Data[S->Top];
}

main.h
函數第二層if分支是遞歸過程

#include <stdio.h>
#include <string.h>
//#include "stack.h"
#define MaxNum 30

int IsFirst = 1;
void f(int A[],int lA,int rA,int B[],int lB,int rB){
	int i;
	if((rA==-1)||(rB==-1)||(rA-lA)!=(rB-lB)) //序列爲空或不一樣長 
	{
		//printf("錯誤序列\n");
		return;
	}
	if(lA==rA&&lB==rB){	//序列都只有一個元素 
		if(A[lA]!=B[lB]){
			//printf("錯誤序列\n");
			return;
		}
		if(IsFirst==1){
			printf("%d",A[lA]);
			IsFirst = 0;
		}
		else{
			printf(" %d",A[lA]);
		}
	}
	else{
		for(i=lB;i<=rB;i++){
			if(A[lA]==B[i]){
				break;
			}
			else if(i==rB){
				//printf("錯誤序列\n");
				return;
			}
		}
		if(i==lB){
			f(A,lA+1,rA,B,lB+1,rB);
			if(IsFirst==1){
				printf("%d",B[i]);
				IsFirst = 0;
			}
			else{
				printf(" %d",B[i]);
			}
		}
		if(i>lB&&i<rB){
			f(A,lA+1,lA+i-lB,B,lB,i-1);
			f(A,lA+i-lB+1,rA,B,i+1,rB);
			if(IsFirst==1){
				printf("%d",B[i]);
				IsFirst = 0;
			}
			else{
				printf(" %d",B[i]);
			}
		}
		if(i==rB){
			f(A,lA+1,rA,B,lB,rB-1);
			if(IsFirst==1){
				printf("%d",B[i]);
				IsFirst = 0;
			}
			else{
				printf(" %d",B[i]);
			}
		}
	}
}

int main(){
	//freopen("1.txt","r",stdin);
	int N,x,i_pre=0,i_in=0;
	char c[5];
	int pre_list[MaxNum];
	int in_list[MaxNum];
	scanf("%d\n",&N);
	Stack s = CreateStack(N);
	for(int i=0;i<2*N;i++){
		scanf("%s",c);
		if(strcmp(c,"Push")==0){
			scanf("%d\n",&x);
			Push(s,x);
			pre_list[i_pre++] = x;
		}
		else{
			in_list[i_in++] = Pop(s);
			getchar();
		}
	}
	f(pre_list,0,N-1,in_list,0,N-1);
	return 0;
}