1086 Tree Traversals Again(树的遍历，

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.

firgure1

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

题目大意

树的递归遍历可以借助堆栈变成非递归遍历，本题提供非递归过程中堆栈操作顺序，以此构建二叉树，并求出这个二叉树的后序。

分析

本题「新瓶装旧酒」，本质依然是考察已知树的两种遍历序列求其他遍历序列，这个问题我已经在一篇博客专门详尽阐述了「知两序求两序」这类问题的解法^[1]，此题只需将问题转化为「知两序求两序」即可解出，观察可知，push的数为前序序列，pop的数为中序序列。

#include <iostream>
#include <vector>
#include <stack>
using namespace std;
vector<int> pre,in,post;
stack<int> st;
void post_travel(int root,int start,int end){
    if(start>end) return;
    int i=start;
    while(i<=end && pre[root]!=in[i]) i++;
    post_travel(root+1,start,i-1);
    post_travel(root+1+i-start,i+1,end);
    post.push_back(pre[root]);
}
int main(){
    int n;
    cin>>n;
    for(int i=0;i<n*2;i++){
        string s;
        int val;
        cin>>s;
        if(s=="Push"){
            cin>>val;
            pre.push_back(val);
            st.push(val);
        }else if(!st.empty()){
            in.push_back(st.top());
            st.pop();
        }
    }
    post_travel(0,0,n-1);
    for(int i=0;i<n;i++){
        if(i==0) cout<<post[i];
        else cout<<" "<<post[i];
    }
    return 0;
}

---

1. https://www.jianshu.com/p/d455cf6aa384 ↩