Tree Traversals Again(根据前序，中序，确定后序顺序）

题目的大意是：进行一系列的操作push,pop。来确定后序遍历的顺序

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.

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 t

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

思路：push过程就是前序遍历，pop过程就是中序遍历，再写一个根据前序和中序遍历确定后序遍历的算法。

 2.取先序序列中的第一个元素，该元素为根结点

    3.根据根结点在中序序列中查找根结点的位置，从而得到该树左子树结点个数（L）与右子树的结点个数（R）

    4.在后序序列数组中，第0到第L个元素为左子树，第L+1到第L+R个元素为右子树，最后一个元素为根结点

 

 

#include<cstdio>
#include<stack>
#include<iostream>
#include<string>
using namespace std;

#define MAX 30
int preOrder[MAX];
int inOrder[MAX];
int postOrder[MAX];


//根据前序和中序划分，来确定后序遍历。前序的第一个数字为根结点，
//找到根结点root在中序数组位置，中序数组中root左边为根结点左子树，右边为右子树
void Solve(int preL,int inL,int postL,int n){   
  if(n==0)return;
  if(n==1){
    postOrder[postL]=preOrder[preL]; 
  }
  int root=preOrder[preL];
  postOrder[postL+n-1]=root;
  int i,R,L;
  for(i=0;i<n;i++){
    if(root==inOrder[inL+i])break;
  }
  L=i,R=n-i-1;                  //L为左子树结点数目，R为右子树结点数目
  Solve(preL+1,inL,postL,L);    //确定后序数组中根结点root左边的排列顺序
  Solve(preL+L+1,inL+L+1,postL+L,R);
}

int main(){
  int n;
  for(int i=0;i<MAX;i++){
    preOrder[i]=0;
    inOrder[i]=0;
    postOrder[i]=0;
  }
  stack<int> s;
  cin>>n;
  string str;
  int data;
  int index=0,pos=0;
  for(int i=0;i<2*n;i++){
    cin>>str;
    if(str=="Push"){      //push代表前序遍历
     cin>>data;
    s.push(data);
      preOrder[index++]=data;
    }else if(str=="Pop"){ //pop为中序遍历
      inOrder[pos++]=s.top();
      s.pop();
    }
  }
  Solve(0,0,0,n);
  for(int i=0;i<n;i++){
    if(i>0)printf(" ");
    printf("%d",postOrder[i]);
  }
  return 0;
}