标签：data mat file each tac ptree and format 一起
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.
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.
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.
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
3 4 2 6 5 1
这个题，我一开始只看到push是按照前序遍历顺序push的，也看到了pop是按照中序遍历顺序pop的，但是没有把二者结合在一起。
蠢到死了啊啊啊。
所以上面说的，就是已经知道前序遍历和中序遍历求后序遍历。
1 /* 2 Push的顺序为前序遍历，Pop的顺序为中序遍历，所以这道题就是 3 已知前序遍历和中序遍历求后序遍历。 4 */ 5 #include <iostream> 6 #include <string> 7 #include <vector> 8 #include <stack> 9 #include <algorithm> 10 using namespace std; 11 12 vector <int> pre, in, post; 13 14 typedef struct Tree { 15 int data; 16 Tree *l, *r; 17 } *ptree; 18 19 ptree build(int pl, int pr, int il, int ir) { 20 if(pl > pr || il > ir) return NULL; 21 int pos; 22 for(int i = il; i <= ir; i ++) { 23 if(in[i] == pre[pl]) { 24 pos = i; 25 break; 26 } 27 } 28 ptree root = new Tree; 29 root -> data = in[pos]; 30 root -> l = root -> r = NULL; 31 root -> l = build(pl + 1, pl + pos - il, il, pos - 1); 32 root -> r = build(pl + pos - il + 1, pr, pos + 1, ir); 33 return root; 34 } 35 36 void post_order(ptree root) { 37 if(!root) return; 38 post_order(root -> l); 39 post_order(root -> r); 40 post.push_back(root -> data); 41 } 42 43 int main() { 44 string op; 45 int n, num; 46 cin >> n; 47 stack <int> s; 48 pre.push_back(-1); 49 in.push_back(-1); 50 for(int i = 0; i < 2 * n; i ++) { 51 cin >> op; 52 if(op == "Push") { 53 cin >> num; 54 pre.push_back(num); 55 s.push(num); 56 } else { 57 in.push_back(s.top()); 58 s.pop(); 59 } 60 } 61 post_order(build(1, n, 1, n)); 62 for(int i = 0; i < post.size(); i ++) { 63 if(i) cout << " "; 64 cout << post[i]; 65 } 66 cout << endl; 67 return 0; 68 }
1086 Tree Traversals Again (25分)(树的重构与遍历)
标签：data mat file each tac ptree and format 一起
原文地址：https://www.cnblogs.com/bianjunting/p/13166569.html
