PAT 1086 Tree Traversals Again[中序转后序][难]
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. Figure 1 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:二叉树的中根遍历,可以通过栈来实现,那么现在给出一棵二叉树的中根遍历操作,要求输出后根遍历结果。 //完全可以通过输入来确定这棵二叉树的中根遍历,即已知中根遍历求后根遍历。但是我不会啊。 ?代码转自: #include <cstdio> #include <vector> #include <stack> #include <cstring> using namespace std; vector<int> pre,in,post,value; void postorder(int root,int start,int end) { if (start > end) return; int i = start; while (i < end && in[i] != pre[root]) {//中序遍历序列中存的节点的id,唯一的! i++; printf("%d %dn",in[i],pre[root]); } postorder(root + 1,start,i - 1); //左子树共有i-start+1个节点。 postorder(root + 1 + i - start,i + 1,end); post.push_back(pre[root]); } int main() { int n; scanf("%d",&n); char str[5]; stack<int> s; int key=0; while (~scanf("%s",str)) { if (strlen(str) == 4) { int num; scanf("%d",&num); value.push_back(num); pre.push_back(key);//对应num有一个序号,从0开始。 s.push(key++); } else { in.push_back(s.top());//现在存了中序遍历 //存的是id对应的序号(为了防止重复呢。) s.pop(); } } postorder(0,0,n - 1); printf("n"); printf("%d",value[post[0]]); for (int i = 1; i < n; i++) printf(" %d",value[post[i]]); return 0; } ? //这个代码简直太难了,看了好几遍都理解不了那个中序转后序的,气死了。 //这个明天还要搜一下别的题解,简直气死我了。 //更要重点掌握一套,二叉树的各种访问序列转换方法。 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |