数据结构课后练习题（练习三）7-5 Tree Traversals Again (25 分)

7-5 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.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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 <iostream>
#include <vector>
#include <stack>
#include <cstring>

using namespace std;

vector<int> pre,in,post,val;

void postorder(int root,int start,int end){
    if(start>end) return ;
    int root_index=0;
    while(root_index<=end&&in[root_index]!=pre[root]) root_index++;
    postorder(root+1,start,root_index-1);
    postorder(root+1+root_index-start,root_index+1,end);
    post.push_back(pre[root]);
}
int main()
{
    stack<int> sta;
    int k=0,N;
    char ch[5];int num;
    scanf("%d",&N);
    while(~scanf("%s",&ch)){
        if(strlen(ch)==4) {
            scanf("%d",&num);
            pre.push_back(num);
            sta.push(num);
        }else{
            in.push_back(sta.top());
            sta.pop();
            if(in.size()==N) break;
        }
    }
    postorder(0,0,N-1);
    for(int i=0;i<N;i++)
        if(i!=N-1) printf("%d ",post[i]);
        else printf("%d",post[i]);

    system("pause");
    return 0;
}

 

标签：Again,25,int,Traversals,tree,pop,Push,Pop,push
来源： https://www.cnblogs.com/littlepage/p/11671482.html