1086. Tree Traversals Again (25) PAT甲级刷题

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

思路：每收到一个push，若下一个操作也为push，则之前那个节点的左孩子为下一节点。相对的每收到一个pop，若下一个操作为push，则为右孩子。用栈来保存信息。

#include 
#include 
#include 
#include 
#include 
using namespace std;

struct TNode{
    int data;
    int left=-1,right=-1;
}tree[31];

int getVal(string op){
    string num;
    for(int i=5;i>ans;
    return ans;
}

void post(int rt){
    static int i = 0;
    if(tree[rt].left!=-1)
        post(tree[rt].left);
    if(tree[rt].right!=-1)
        post(tree[rt].right);
    if(i++==0)
        cout<>n;
    char c = cin.get();
    vector ops;
    for(int i=0;i s;
    for(int i=0;i