技術標籤：PAT甲級簡單題二叉樹

A 1086 Tree Traversals Again

題目連結

Problem Description

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.

Input

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

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

題目大意：

給出一段棧模擬命令，出棧順序就是一顆二叉樹的中序序列。輸出這顆二叉樹的後序序列。

解題思路：

得到隱藏條件——入棧順序是一顆二叉樹的先序序列，然後根據這兩個序列找到所有根節點及其對應的左右子樹，將根節點放在左右子樹後面輸出，就得到後序序列了。

AC程式碼

#include<bits/stdc++.h>
using namespace std;
const int maxn = 35;

int in[maxn],pre[maxn],n,m,node,ind=0,pnd=0,cnt=0;
string str;
stack<int>stk;

void dfs(int inL,int inR,int preL,int preR){
    if(preL>preR) return;
    int root=pre[preL],k;
    for(;k<inR;k++){
        if(in[k]==root) break;
    }
    int leftnum=k-inL;
    dfs(inL,k-1,preL+1,preL+leftnum);
    dfs(k+1,inR,preL+leftnum+1,preR);
    if(cnt==0){
        printf("%d",root);
        cnt++;
    }else{
        printf(" %d",root);
        cnt++;
    }
    return;
}

int main(){
    scanf("%d",&n);
    m=2*n;
    while(m--){
        cin>>str;
        if(str=="Push"){
            cin>>node;
            stk.push(node);
            pre[pnd++]=node;
        }else{
            in[ind++]=stk.top();
            stk.pop();
        }
    }
    dfs(0,n-1,0,n-1);
    return 0;
}