1086 Tree Traversals Again (PAT甲级)

#include 
#include 
#include 
#include 
#include 

int N, t;
std::string str;
std::stack vec;
std::vector preorder, inorder, postorder;
std::map mp;

void construct(int preL, int preR, int inL, int inR){
    if(preL == preR){
        return;
    }
    int loc = mp[preorder[preL]];
    construct(preL + 1, loc - inL + preL + 1, inL, loc);
    construct(loc - inL + preL + 1, preR, loc + 1, inR);
    postorder.push_back(preorder[preL]);
}

int main(){
    std::cin >> N;
    for(int i = 0; i < 2 * N; ++i){
        std::cin >> str;
        if(str == "Push"){
            std::cin >> t;
            preorder.push_back(t);
            vec.push(t);
        } else{
            inorder.push_back(vec.top());
            vec.pop();
        }
    }
    for(int i = 0; i < N; ++i){
        mp[inorder[i]] = i;
    }
    construct(0, N, 0, N);
    for(int i = 0; i < postorder.size(); ++i){
        printf("%d%s", postorder[i], i == postorder.size() - 1 ? "\n" : " ");
    }
    return 0;
}

题目如下:

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.

1086 Tree Traversals Again (PAT甲级)_第1张图片


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 2Nlines 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

 

你可能感兴趣的:(PAT甲级,pat考试)