03-树3 Tree Traversals Again (25分)

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 NNN (≤30\le 30≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to NNN). Then 2N2N2N 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 <iostream>
#include <cstdio>
#include <stack>
#include <cstring>
using namespace std;
typedef int ElementType;
typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
    ElementType Data;
    BinTree Left;
    BinTree Right;
};
int post[30], pre[30], in[30];
//递归思想解决，前三个参数分表表示要解决的那个递归中数组开始的下标
void pre_in_to_post(int preLoc, int inLoc, int postLoc, int n){
    if(n == 0)  return;         //右边或左边没有元素的时候
    if(n == 1)  post[postLoc] = pre[preLoc];    //如果左边或右边只剩一个元素则pre数组中唯一的一个元素就是post数组中的元素
    int L, R, root;             //L，R表示左右数组的规模，
    root = pre[preLoc];
    post[postLoc + n - 1] = root;
    int i;
    for(i = 0; i < n; i++)
        if(root == in[inLoc + i])
            break;
    L = i;R = n - L - 1;
    pre_in_to_post(preLoc + 1, inLoc, postLoc,L);
    pre_in_to_post(preLoc+L + 1, inLoc+L + 1,postLoc + L,R);
}
int main(){
    int N,i = 1,j = 1;
    char op[7];     //读取一行的输入再处理
    char op_[4];
    stack<int> s;
    scanf("%d", &N);
    getchar();      //吸收回车，以防被下面gets读取
    for(int k = 0;k < 2*N; k++){
        gets(op);
        if(op[1] == 'u'){
            sscanf(op, "%s %d", op_, &pre[j]);//将一行分为一个字符串和一个整数
            s.push(pre[j]);
            j++;
        }else{
            int temp = s.top();
            in[i++] = temp;
            s.pop();
        }
    }

    pre_in_to_post(0, 0, 0, i);

    for(int k = 0;k < N; k++)
        if(k != 0) putchar(' ');
        printf("%d", post[k]);
    return 0;
}