1086. Tree Traversals Again (25)-PAT甲级真题

1086. Tree Traversals Again (25)
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：用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历
分析：栈实现的是二叉树的中序遍历（左根右），而每次push入值的顺序是二叉树的前序遍历（根左右），所以该题可以用二叉树前序和中序转后序的方法做~~

root为当前子树的根结点在前序pre中的下标，start和end为当前子树的最左边和最右边的结点在中序in中的下标。用i找到当前子树的根结点root在中序中的下标，然后左边和右边就分别为当前根结点root的左子树和右子树。递归实现~

#include <cstdio>#include <vector>#include <stack>#include <cstring>using namespace std;vector<int> pre, in, post;void postorder(int root, int start, int end) {    if(start > end) return ;    int i = start;    while(i < end && in[i] != pre[root]) i++;    postorder(root + 1, start, i - 1);    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;    while(~scanf("%s", str)) {        if(strlen(str) == 4) {            int num;            scanf("%d", &num);            pre.push_back(num);            s.push(num);        } else {            in.push_back(s.top());            s.pop();        }    }    postorder(0, 0, n - 1);    printf("%d", post[0]);    for(int i = 1; i < n; i++)        printf(" %d", post[i]);    return 0;}