2020-03-30

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.

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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

#include
#include
#include
#include
using namespace std;
vector pre,mid,back;
void postorder(int root,int start,int end){
    if(start>end){
        return;
    }
    int i = start;
    while(i stack;
    cin>>node;
    string s;
    int m;
    for(int i=0;i<2*node;++i){
        cin>>s;
        if(s[1] == 'u'){
            cin>>m;
            pre.push_back(m);
            stack.push(m);
        }else{
            mid.push_back(stack.top());
            stack.pop();
        }
    }
    postorder(0,0,node-1);
    printf("%d",back[0]);
    for(int i=1;i