标签:1086 PAT 25 int tree Pop push pop Push
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.
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:
给出一个二叉树栈实现的中序遍历执行过程,求它的后序遍历序列。
题目解析:
中序序列可以由栈操作得到,而数据的输入正好是先序序列。在中序序列里面找到root的位置,左边是左子树,右边是右子树 。
具体代码:
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
#define MAXN 50
int pre[MAXN],in[MAXN];
stack<int> s;
void post(int root,int start,int end){
if(start>end) return;
int i=start;
while(i<end&&in[i]!=pre[root]){i++;}//在中序序列里面找到root的位置,左边是左子树,右边是右子树
// cout<<pre[root]<<" "<<in[i]<<" "<<pre[root+1]<<" "<<pre[root+i-start+1]<<endl;
post(root+1,start,i-1);
post(root+i-start+1,i+1,end);
if(root!=0)
printf("%d ",pre[root]);
else
printf("%d",pre[root]);
}
int main()
{
int n,k=0,k0=0;
cin>>n;
for(int i=0;i<2*n;i++){
string str;
cin>>str;
if(str=="Push"){
cin>>pre[k];
s.push(pre[k]);
k++;
}else{
in[k0]=s.top();
s.pop();
k0++;
}
}
post(0,0,n-1);
return 0;
}
标签:1086,PAT,25,int,tree,Pop,push,pop,Push 来源: https://blog.csdn.net/qq_29978597/article/details/86571957
本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享; 2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关; 3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关; 4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除; 5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。