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<cstdio>

 #include<iostream>

 #include<stack>

 #include<string.h>

 using namespace std;

 typedef struct NODE{

     NODE* left, *right;

     int data;

 }node;

 stack<int> stk;

 int pre[], in[], N;

 node* create(int preL, int preR, int inL, int inR){

     if(preL > preR){

         return NULL;

     }

     node *root = new node;

     root->data = pre[preL];

     int i;

     for(i = inL; i <= inR; i++)

         if(in[i] == root->data)

             break;

     int Lnum = i - inL;

     root->left = create(preL + , preL + Lnum, inL, i - );

     root->right = create(preL + Lnum + , preR, i + , inR);

     return root;

 }

 void post(node *tree, int &cnt){

     if(tree == NULL)

         return;

     post(tree->left, cnt);

     post(tree->right, cnt);

     if(cnt == N - )

         printf("%d", tree->data);

     else{

         printf("%d ", tree->data);

         cnt++;

     }

 }

 int main(){

     int num, indexPre = , indexIn = ;

     char str[];

     scanf("%d", &N);

     for(int i = ; i < *N; i++){

         scanf("%s", str);

         if(strcmp(str, "Push") == ){

             scanf("%d ", &num);

             stk.push(num);

             pre[indexPre++] = num;

         }else{

             num = stk.top();

             stk.pop();

             in[indexIn++] = num;

         }

     }

     node *tree = create(, N - , , N - );

     int cnt = ;

     post(tree, cnt);

     cin >> N;

     return ;

 }

总结：

1、中序遍历的非递归实现：不断将非空的左孩子入栈，当左边为空时，弹出一个栈顶元素访问之，并将指针移至他的右子树，继续进行开始时的操作。 在这个过程中，push的特点是不断把遇到的新节点push入栈，因此push的过程就是先序遍历的过程。而pop自然是中序的过程。因此，中序与先序的非递归实现的区别就在于访问节点的时机不同，先序在push时，中序在pop时。

2、因此可以根据题目同时建立一个栈，同步做push和pop操作，先得到先序与中序遍历序列，再按照套路建树。