pat 1086. Tree Traversals Again (25)

pat 1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

// 这道题目，大家想清楚了，就直接用push,pop 来做吧，然后很容易就得出结果了。记得最大的数字是2*n。不过思考的过程还是需要的，过程有点麻烦。

//
//  pat1086.cpp
//  acmProject
//
//  Created by HuQiaoNan on 14-9-6.
//  Copyright (c) 2014年 HuQiaoNan. All rights reserved.

#include <stdio.h>

struct Node
{
    int val;
    Node *left;
    Node *right;
    Node(int v):val(v),left(NULL),right(NULL){}
    Node(){}
};
char input[8];
int n=0,ids=0,ids2=0;
Node * buildTree()
{
    Node * r=NULL;
    int tmp=0;
    if(ids<2*n)
    {
        scanf("%s",input);
        if(input[1]=='u')
        {
            scanf("%d",&tmp);
            r = new Node(tmp);
            ids++;
        }else if(input[1]=='o')
        {
            ids++;
            return NULL;
        }
       r->left = buildTree();
        r->right = buildTree();
    }
    return r;
}
void postorder(Node *r)
{
    if(r!=NULL)
    {
        postorder(r->left);
        postorder(r->right);
        printf("%d",r->val);
        ids2++;
        if(ids2<n){
            printf(" ");
        }else
        {
            printf("\n");
        }
    }
}
int main()
{
    scanf("%d",&n);
    Node * root = buildTree();
    postorder(root);
    return 0;
}