03-树3 Tree Traversals Again （c++递归实现）-蒲公英云

03-树3 Tree Traversals Again （c++递归实现）

题目

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 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

题意理解

push的顺序为先序遍历
pop的顺序为中序遍历
按照后序遍历输出

代码

#include<cstdio>
#include<stack>
#include<string>
#include<iostream>
using namespace std;
const int maxn=30;
int pre[maxn],in[maxn],post[maxn];    //    分别代表前序，中序，后序遍历数组
//PreL 前序遍历数组的位置
//inL 中序遍历数组的位置
//postL 后序遍历数组的位置
void solve(int PreL, int inL, int postL, int n){
    if(n==0) return;
    if(n==1){
        post[postL]=pre[PreL];
        return;
    }
    int i;
    int root=pre[PreL];            //前序遍历数组的第PreL个数就是根结点
    post[postL+n-1]=root;        //root是后序遍历数组的第postL+n-1个数
    for(i=0;i<n;i++){
        if(in[inL+i]==root) break;    //在中序遍历数组中找到root,记录root所在位置i
    }    
    int L=i,R=n-i-1;            //root左边的是左子树，右边的是右子树
    solve(PreL+1,inL,postL,L);    //递归解决左子树
    solve(PreL+1+L,inL+L+1,postL+L,R);    //递归解决右子树
}
int main(){
    int n,num,i=0,j=0;
    scanf("%d",&n);                //输入一个整数
    string s;
    stack<int> st;
    for(int z=0;z<2*n;z++){        //读入2n行，构建前序遍历和中序遍历的数组
        cin>>s;                    //读入字符串判断入栈或出栈    
        if(s=="Push"){            //入栈
            scanf("%d\n",&num); //输入要入栈的数字num
            st.push(num);        //num入栈
            pre[i++]=num;        //把num放入前序遍历数组
        }else{                    //出栈
            num=st.top();            //取出栈顶赋给num
            st.pop();            //出栈
            in[j++]=num;        //num赋给中序遍历数组
        }
    }
    solve(0,0,0,n);                //通过前序遍历和中序遍历求二叉树的后序遍历
    for(i=0;i<n;i++){
        printf("%d",post[i]);
        if(i!=n-1) printf(" ");
    }
    return 0;
}