1086. 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.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer

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


push的顺序实际上是前序遍历序列，而pop的顺序实际上是中序遍历序列，通过这两个序列可以确定树，但比较麻烦。所以可以根据中序遍历的特点来建树。pop之后的push是右子树结点，连续的push是左子树结点，所以可以设立一个变量status来判断是左子树还是右子树，连续的pop中最后一次pop先不弹栈，因为接下来的push是根结点的右子树。
代码：

#include <stdio.h>
#include <string.h>
struct tree {
    int left,right;
}s[31];
int flag = 0;
void postorder(int t) {
    if(t == -1)return;
    postorder(s[t].left);
    postorder(s[t].right);
    if(!flag) {
        flag = 1;
        printf("%d",t);
    }
    else printf(" %d",t);
}
int sta[31],no = 0;
int main() {
    char op[5];
    int n,d,root = -1;
    scanf("%d",&n);
    for(int i = 0;i < n;i ++) {
        s[i + 1].left = s[i + 1].right = -1;
    }
    int status = 0;///代表状态，0表示是树根，1表示是栈顶元素的左子树，2表示是栈顶元素的右子树，这个时候要将栈顶元素弹出。
    for(int i = 0;i < n * 2;i ++) {
        scanf("%s",op);
        if(strcmp(op,"Push") == 0) {
            scanf("%d",&d);
            if(status == 1) {
                s[sta[no - 1]].left = d;
                sta[no ++] = d;//这里不是no ++，因为栈顶元素是早就该出栈却特意保留的。
            }
            else if(status == 2) {
                s[sta[no - 1]].right = d;
                sta[no - 1] = d;
                status = 1;
            }
            else {
                root = d;
                status = 1;
                sta[no ++] = d;
            }
        }
        else {
            if(status == 1) status = 2;
            else no --;
        }
    }
    postorder(root);
}