1086 Tree Traversals Again （25 分)

2019-04-10

1086 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 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:

1	3 4 2 6 5 1

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出中序遍历一棵树的过程，求这棵树的后序遍历

分析

一开始没想明白。其实push的顺序就是前序遍历，pop的顺序是中序，知道前序中序，后序就好办了

代码

#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include <sstream>

using namespace std;
int n;
bool flag=true;
vector<int> pre;
vector<int> in;
stack<int> s;
void postorder(int root,int start,int end){
    if(start>end)
        return;
    int i = start;
    while(i<end && in[i] != pre[root] ) i++;
    postorder(root+1,start,i-1);
    postorder(root+1+i-start,i+1,end);
    printf("%s%d",flag==true?"":" ",pre[root]);
    flag = false;
}

int main() {
    cin>>n;
    int x;
    string str;
    for(int i=0;i<2*n;i++){
        cin>>str;
        if(str.length()==4){
            cin>>x;
            pre.push_back(x);
            s.push(x);
        }else{
            in.push_back(s.top());
            s.pop();
        }
    }
    postorder(0,0,n-1);
    return 0;
}

其他

读取数据的时候，如果cin和getline混用最后一个测试点会段错误，很奇怪，不知道为什么。