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  • 由先序和中序求后序

    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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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
    题意:按先序输入,pop出的顺序是中序,输出后序。
     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<cstring>
     4 #include<iostream>
     5 #include<stack>
     6 using namespace std;
     7 struct TNode
     8 {
     9     int left;
    10     int right;
    11 } Tree[35];//树的节点最多不超过 30
    12 int s;//用来保存树的根的值  因为好像第一个测试用例的树根不是  1
    13 void Create()
    14 {  memset(Tree,0,sizeof(Tree));
    15     int n;
    16     scanf("%d",&n);
    17     cin.get();
    18     stack<int >S;
    19     int top;
    20     char str[10];
    21     int data;
    22     scanf("%s %d
    ",str,&data);
    23     S.push(data);
    24     s=top=S.top();
    25   
    26     for (int i=1; i<2*n; i++)//树根已入栈,故从i=1开始
    27     {
    28         scanf("%s",str);
    29         if(strlen(str)!=3)//注意这个if结构
    30         {
    31             scanf("%d",&data);
    32             S.push(data);
    33             if(Tree[top].left==0)//左子树空,将data放入左子树
    34                 Tree[top].left=data;
    35             else //左子树已被使用
    36                 Tree[top].right=data;
    37             top=S.top();//更新top
    38         }
    39         else
    40         {//注意这个大括号里的代码
    41             top=S.top();
    42             S.pop();
    43         }
    44     }
    45 }
    46 int k=0;
    47 void Merge(  int n)//控制输入输出格式
    48 {
    49 
    50     if(Tree[n].left!=0)
    51         Merge(Tree[n].left);
    52     if(Tree[n].right!=0)
    53         Merge(Tree[n].right);
    54     if(k==0)
    55     {
    56         printf("%d",n);
    57         k=1;
    58     }
    59     else
    60         printf(" %d",n);
    61 }
    62 int main()
    63 {
    64     Create();
    65     Merge(s);
    66     return 0;
    67 
    }
    
    
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  • 原文地址:https://www.cnblogs.com/hbhdhd/p/10859090.html
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