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

一箇中序二叉樹遍歷可以用非遞迴方式用棧來實現。比如，假設遍歷一個有6個節點的二叉樹（鍵值從1-6），棧的操作如下：push 1;push 2;push 3;pop();pop();push 4; pop();pop(); push 5; push 6; pop(); pop(). 然後通過這一系列操作就可以構造出如圖1的二叉樹，你的任務是給出這棵樹的後序遍歷

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.

第一行包含一個正整數N，表示這棵樹的總節點數（節點從1-N進行編號）。之後有2Ｎ行，每一行描述了一個棧操作 

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.

打印出正確的二叉樹的後序遍歷序列，所有數字用空格隔開，最後一個數字之後沒有空格 

思路：通過棧操作構造二叉樹再進行後序遍歷。

　　　Push的次序就是先序遍歷序列中元素的順序。Pop則是中序遍歷序列中元素的順序。 

          即知道先序遍歷和中序遍歷求後序遍歷。

#include<cstdio>
#include<cstring>
#include<stack>
#include<algorithm>
using namespace std;
const int maxn = 50;

struct node{
	int data;
	node* lchild;
	node* rchild;
};
int pre[maxn],in[maxn],post[maxn];
int n;
//------根據前序遍歷和中序遍歷構造二叉樹------ 
node* create(int preL,int preR,int inL,int inR){
	if(preL>preR) return NULL;
	node* root = new node;
	root ->data = pre[preL];
	
	int k;
	for(k=inL;k<inR;k++){
		if(in[k]==pre[preL]){
			break;
		}
	}
	
	int numLeft = k-inL;
	root->lchild=create(preL+1,preL+numLeft,inL,k-1);
	root->rchild=create(preL+numLeft+1,preR,k+1,inR);
	return root;
} 

int num=0;
//-------後序遍歷---------- 
void postorder(node* root){
	if(root== NULL) return;
	postorder(root->lchild);
	postorder(root->rchild);
	printf("%d",root->data);
	num++;
	if(num<n) printf(" ");
}

int main(){
	scanf("%d",&n);
	char str[5];
	stack<int> st;
	int x,preIndex=0,inIndex=0;
	
	for(int i=0;i<2*n;i++){
		scanf("%s",str);
		if(strcmp(str,"Push")==0){//如果是Push,入棧，放入前序遍歷陣列 
			scanf("%d",&x);
			pre[preIndex++]=x;
			st.push(x);
		}else{//如果是Pop 出棧，放入中序遍歷陣列 
			in[inIndex++]=st.top();
			st.pop();
		}
	}
	node* root = create(0,n-1,0,n-1);
	postorder(root);
	return 0;
}