PAT (Advanced Level) Practise 1086 Tree Traversals Again (25)

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本文介绍了一种通过栈操作序列构建唯一二叉树的方法，并给出了实现该过程的C++代码示例。任务包括解析输入的栈操作序列，构建对应的二叉树并输出其后序遍历序列。

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1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意：通过栈操作构建出一棵树，并输出它的后序遍历

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n,x,a[35][2],flag;
char ch[10];

void dfs(int k)
{
    if(a[k][0]!=-1) dfs(a[k][0]);
    if(a[k][1]!=-1) dfs(a[k][1]);
    if(flag) printf("%d",k);
    else printf(" %d",k);
    flag=0;
}

int main()
{
    while(~scanf("%d",&n))
    {
        memset(a,-1,sizeof a);
        stack<int>s;
        int root=-1,pre;
        for(int i=1;i<=n*2;i++)
        {
            scanf("%s",ch);
            if(!strcmp(ch,"Push"))
            {
                scanf("%d",&x);
                if(root==-1) {root=x;s.push(x);continue;}
                if(!s.empty()&&a[s.top()][0]==-1) a[s.top()][0]=x;
                else a[pre][1]=x;
                s.push(x);
            }
            else
            {
                pre=s.top();
                s.pop();
            }
        }
        flag=1;
        dfs(root);
        printf("\n");
    }
    return 0;
}