PTA 03-树3 Tree Traversals Again (25分)

题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/667

5-5 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 NN (\le 30≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to NN). Then 2N2Nlines 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



这道题是根据输入算出前序遍历和中序遍历，不过没有直接建树去做后序，而是参照前两者遍历的序列，递归分解算了后续出来

/*
评测结果
时间	结果	得分	题目	编译器	用时（ms）	内存（MB）	用户
2017-07-08 16:00	答案正确	25	5-5	gcc	14	1
测试点结果
测试点	结果	得分/满分	用时（ms）	内存（MB）
测试点1	答案正确	12/12	14	1
测试点2	答案正确	4/4	2	1
测试点3	答案正确	4/4	2	1
测试点4	答案正确	1/1	13	1
测试点5	答案正确	4/4	2	1

*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAXLEN 50
struct stack{
	int data[MAXLEN];
	int top;
};

typedef struct stack *ptrStack;

int printcount=0;
int length;

ptrStack CreateStack()
{
	ptrStack temp;
	temp=(ptrStack)malloc(sizeof(struct stack));
	temp->top=0;
	return temp;
}

void Push(ptrStack s,int item)
{
	s->data[++(s->top)]=item;
}

int Pop(ptrStack s)
{
	return s->data[(s->top)--];
}

void DestroyStack(ptrStack s)
{
	free(s);
}

void Process(int preOrder[],int inOrder[],int preStart,int preEnd,int inStart,int inEnd)
{
	int i,root,mid,leftsize,rightsize;
	int nextLeftpreStart,nextLeftpreEnd,nextLeftinStart,nextLeftinEnd;
	int nextRightpreStart,nextRightpreEnd,nextRightinStart,nextRightinEnd;
	if(preStart==preEnd){
		printf("%d",preOrder[preStart]);
		printcount++;
		if (printcount!=length)
			putchar(' ');
		return;
	}
	root=preOrder[preStart];
	for(i=inStart;i<=inEnd;i++)
		if(inOrder[i]==root)
			break;
	leftsize=i-inStart;
	rightsize=inEnd-i;

	nextLeftpreStart=preStart+1;
	nextLeftpreEnd=preStart+leftsize;
	nextRightpreStart=nextLeftpreEnd+1;
	nextRightpreEnd=nextLeftpreEnd+rightsize;

	nextLeftinStart=i-leftsize;
	nextLeftinEnd=i-1;
	nextRightinStart=i+1;
	nextRightinEnd=i+rightsize;

	if(i!=inStart){
		Process(preOrder,inOrder,nextLeftpreStart,nextLeftpreEnd,nextLeftinStart,nextLeftinEnd);
	}

	if(i!=inEnd){
		Process(preOrder,inOrder,nextRightpreStart,nextRightpreEnd,nextRightinStart,nextRightinEnd);
	}
	printf("%d",root);
	printcount++;
	if (printcount!=length)
		putchar(' ');
	return;
}

void Input(int pre[],int in[])
{
	int i,n,num,idxforpush=0,idxforpop=0;
	char* s;
	char temp[50];
	ptrStack workstack=CreateStack();

	scanf("%d",&n);
	length=n;
	n=n*2;
	for(i=0;i<n;i++)
	{

		scanf("%s",temp);
		s=strchr(temp,'h');
		if(s!=NULL){
			scanf("%d",&num);
			pre[idxforpush]=num;
			Push(workstack,num);
//			printf("----pre----%d\n",pre[idxforpush]);//for test
			idxforpush++;

		}
		else{
			in[idxforpop]=Pop(workstack);
//			printf("++++ in++++%d\n",in[idxforpop]);//for test
			idxforpop++;
		}

	}
	DestroyStack(workstack);
}
int main()
{
	int i;
	int in[MAXLEN];
	int pre[MAXLEN];
	Input(pre,in);
//	for(i=0;i<length;i++)		printf("idx %d   in %d  pre %d\n",i,in[i],pre[i]);//for test;
	Process(pre,in,0,length-1,0,length-1);
}