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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 PopSample Output:
3 4 2 6 5 1
#include <stdio.h> #define MaxSize 35 #define EMPTY_TOS (-1) #define PUSH 1 #define POP 2 typedef struct stack_record { int top_of_stack; int stack_array[MaxSize]; } Stack; void init_stack(Stack *S) { S->top_of_stack=EMPTY_TOS; } int is_empty(Stack *S) { return (S->top_of_stack==EMPTY_TOS); } int is_full(Stack *S) { return (S->top_of_stack==MaxSize-1); } void push(int x,Stack *S) { if(is_full(S)) return; else S->stack_array[++S->top_of_stack]=x; } int top(Stack *S){ if(!is_empty(S)) return S->stack_array[S->top_of_stack]; } int pop(Stack *S) { if(is_empty(S)) return; else return S->stack_array[S->top_of_stack--]; } int main () { Stack S1,S2; int OpArray[2*MaxSize+1][2]={-1}; char tmpstr[30],tmp[30],tmp1[30]; init_stack(&S1); init_stack(&S2); int i,NodeNum,tmppop,tmppop1; int PrintTag=0; scanf("%d",&NodeNum); getchar(); for (i=0;i<2*NodeNum;i++){ gets(tmpstr); if(tmpstr[1]==‘u‘){ sscanf(tmpstr,"%s%s",tmp,tmp1); OpArray[i][0]=PUSH; OpArray[i][1]=atoi(tmp1); }else { OpArray[i][0]=POP; } } for (i=0;i<2*NodeNum;i++){ if(OpArray[i][0]==PUSH) push(OpArray[i][1],&S1); else{ if(OpArray[i+1][0]==PUSH){ if(top(&S1)==-1){ push(OpArray[i+1][1],&S2); }else{ push(-1,&S2); push(pop(&S1),&S2); push(-1,&S1); push(OpArray[i+1][1],&S2); } i++; } else{ tmppop=pop(&S1); if(tmppop==-1){ while((tmppop1=pop(&S2))!=-1) printf(" %d",tmppop1); }else{ if (!PrintTag){ printf("%d",tmppop); PrintTag++; }else printf(" %d",tmppop); } } } } return 0; }
03-3. Tree Traversals Again (25)
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原文地址：http://www.cnblogs.com/RbtreeLinux/p/4178628.html
