1086.Tree Traversals Again

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.

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

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结题思路：

根据输入，构建二叉树。
每当上一个读入的OP为Push，或者第一次读入Push时，当前的Push操作在当前节点的左子树插值
每当上一个命令的OP为Pop ，当前的Push操作在当前节点的右子树插值
建树完成后，后序遍历输出即可。
需要注意：根节点的索引并不确定。考虑空树跟只有一个节点的情况

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cstdlib>
#include<stack>
#define size 35
using namespace std;
struct node{
       int left;
       int right;
};
stack<int> s;
node tree[size];//数组存储树 
void addChild(int root,int child,bool flag)
{
     if(child==root)
      return;
     if(flag)//left
     {
             tree[root].left=child;
     }
     else
     {
             tree[root].right=child;
     }
}
void postOrder(int root,int first)
{
     if(root==-1)
        return ;
     postOrder(tree[root].left,first);
     postOrder(tree[root].right,first);
     if(root==first)
     cout<<root<<endl;
     else
     cout<<root<<" ";
}
void postOrderOutTree(int root)
{
     if(root==0||root==-1)
        return ;  
     postOrder(root,root);
}
int main()
{   
    string op="";//操作符 
    int cur;//当前节点的索引值 
    int nodeNum;//树节点个数 
    int nodeValue;//插入的树节点的值 
    bool flag;//左右子树的标识符 
    int root;//根节点的位置 
    bool first=true;//根节点的标示符 
    while(!s.empty())//栈初始化 
           s.pop();
    for(int i=0;i<size;++i)//数组存储树初始化
        tree[i].left=tree[i].right=-1;
    cin>>nodeNum;
    cur=0;
    root=0;
    flag=true;//在0节点的左子树处插入 
    for(int i=0;i<nodeNum*2;++i)//建树 
    {
        cin>>op;
        if(op=="Push")//节点入栈 
        {
           cin>>nodeValue;
           if(first)//记录根节点的索引值 
           {
              root=nodeValue;
              first=false;
           }
           s.push(nodeValue);
           addChild(cur,nodeValue,flag);//flag 1 左子树 0 右子树，给cur节点添加孩子nodeValue 
           cur=nodeValue;//修改当前节点的索引值 
           flag=true;
        }
        else
        {
            cur=s.top();//修改当前节点的索引值 
            s.pop();
            flag=false;//在当前节点的右子树插入 
        }
    } 
    postOrderOutTree(root);//后序遍历，并输出 
    //system("pause");
    return 0;
}