1086. Tree Traversals Again (25)

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 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


/*************************************************************
Author : qmeng
MailTo : qmeng1128@163.com
QQ     : 1163306125
Blog   : http://blog.youkuaiyun.com/Mq_Go/
Create : 2018-03-01 16:22:45
Version: 1.0
**************************************************************/
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <stack>
#include <cstring>
using namespace std;
struct Tree{
    Tree *li;
    Tree *ri;
    int data;
};
Tree *Tr;
stack<Tree *> q;
int k = 1;
int print(Tree *x){
    if (x == NULL)return 0;
    print(x->li);
    print(x->ri);
    if(k==0){
        printf(" %d",x->data);
    }else{
        printf("%d",x->data);
        k--;
    }
    return 0;
}
int main(){
    int n;
    int temp;
    scanf("%d",&n);
    string str;
    cin >> str;
    cin >> temp;
    Tree *Tr = (Tree *)malloc(sizeof(Tree));
    Tr->data = temp;
    Tr->li = NULL;
    Tr->ri = NULL;
    q.push(Tr);

    Tree *root = q.top();
    int kk = 0;
    while(kk < n){
        cin >> str;
        if(str=="Push"){
            cin >> temp;
            Tree *t = (Tree *)malloc(sizeof(Tree));
            t->data = temp;
            t->li = NULL;
            t->ri = NULL;
            if(root->li == NULL){
                root->li = t;
            }else{
                root->ri = t;
            }
            q.push(t);
            root = t;
        }else{
            kk++;
            root = q.top();
            q.pop();

        }
    }
    print(Tr);
    return 0;
}
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