• 03-3. Tree Traversals Again (25)


    03-3. Tree Traversals Again (25)

    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
    

    #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;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/RbtreeLinux/p/4178628.html
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