标签:1086 PAT Again int tree pop Push Pop push
https://pintia.cn/problem-sets/994805342720868352/problems/994805380754817024
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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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
代码:
#include <bits/stdc++.h> using namespace std; int N; vector<int> in, post, pre, val; void postorder(int root, int st, int en) { if(st > en) return; int i = st; while(i < en && in[i] != pre[root]) i ++; postorder(root + 1, st, i - 1); postorder(root + 1 + i - st, i + 1, en); post.push_back(pre[root]); } int main() { scanf("%d", &N); stack<int> s; string op; int cnt = 0; for(int t = 0; t < N * 2; t ++) { cin >> op; if(op == "Push") { int x; scanf("%d", &x); pre.push_back(cnt); val.push_back(x); s.push(cnt ++); } else { in.push_back(s.top()); s.pop(); } } postorder(0, 0, N - 1); for(int i = 0; i < N; i ++) { printf("%d", val[post[i]]); printf("%s", i != N - 1 ? " " : ""); } return 0; }
push 的顺序是前序遍历的顺序 按照题目 pop 得到的中序遍历的顺便 in 和 pre 存的是数字的位置 val 求数字的值 递归求出后序遍历
FH
标签:1086,PAT,Again,int,tree,pop,Push,Pop,push 来源: https://www.cnblogs.com/zlrrrr/p/10364656.html
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。