Tree Traversals Again

题目地址http://www.patest.cn/contests/mooc-ds2015spring/03-%E6%A0%912

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

 解题思路：主要就是找规律建树

代码如下

#include <iostream>
#include <string>
#include <vector>
using namespace std;
typedef struct  TreeNode  *Bintree;
struct  TreeNode{
    int Data;
    Bintree Left;
    Bintree Right;
};
int flag = 0;
void PreOrderTraversal(Bintree BT)
{
    if (BT)
    {
        PreOrderTraversal(BT->Left);
        PreOrderTraversal(BT->Right);
        if (flag == 0)
        {
            cout << BT->Data;
            flag = 1;
        }
        else 
        cout << " "<<BT->Data;
    }
}
TreeNode node[31];

int main(){
    int number;
    cin >> number;
    for (int i = 1; i <= number;i++)
    {
        node[i].Data = i;
    }
    int root;
    string  opertion;
    string  lastOpertion;
    int  nodeNum = 0;
    vector<int>  stacktree;
    int lastNodeNum = 0;
    cin >> opertion;
    if (opertion == "Push")
    {
        cin >> nodeNum;
        stacktree.push_back(nodeNum);
    }
    root = nodeNum;
    lastOpertion = opertion;
    for (int i = 0; i < 2 * number-1; i++)
    {
        cin >> opertion;
        if (opertion == "Push")
        {
            cin >> nodeNum;
            if (lastOpertion == "Push")
            {
                node[stacktree[stacktree.size() - 1]].Left = &node[nodeNum];
            }
            else
            {
                node[lastNodeNum].Right = &node[nodeNum];
            }
            stacktree.push_back(nodeNum);
        }
        if (opertion == "Pop")
        {
            lastNodeNum = stacktree[stacktree.size()-1];
            stacktree.pop_back();
        }
        lastOpertion = opertion;
    }
    PreOrderTraversal(&node[root]);

    return 0;
}