【树】7-5Tree Traversals Again

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 (≤) 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.

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：

• 通过观察可知入栈顺序即先序遍历，出栈顺序即中序遍历，所以在得知先序和中序遍历的基础上还原树，然后对该树进行后序遍历。
• 其中，为了得到先序（中序）结果，对每一条命令进行判断，如果第二个字符是“u”说明是“push”，然后从第五个位置开始截取就可获得数字，同时将该数字入栈；否则是“pop”，将栈中的数字出栈。
• 在得到先序和中序序列之后，从前往后遍历先序序列，每次从中序序列中找到左右子树的范围，使用left和right标记，进行递归，如果left！=right，说明当前结点不是叶子结点，继续递归左右子树；如果left==right，说明该结点是叶子结点。
• 注意，其中的两个变量cur和printtag必须设成全局变量，因为使用该变量的函数中有递归函数，如果cur设成局部变量会发生段错误；如果printtag设成局部变量，会发生输出格式错误。

解题思路和代码参考https://blog.csdn.net/xyt8023y/article/details/47443489

#include <iostream>
#include <cstdlib>
#include <vector>
#include <stack>
#include <string>
#include <sstream>
using namespace std;

//分别存储先序序列和中序序列
vector<int>preorder;
vector<int>inorder;

typedef struct TreeNode* Node;
struct TreeNode
{
    int num;
    Node left;
    Node right;
};

//根据结点值在中序序列中寻找根结点的下标位置
int FindRootIndex(int rootnum) {
    for (unsigned int i = 0; i < inorder.size(); i++) {
        if (inorder[i] == rootnum)
            return i;
    }
    return -1;
}

//通过给出中序序列中左右子树的范围建树
int cur;//定义遍历preorder的变量
Node CreateTree(int left, int right) {
    //int cur = 0;
    if (left > right)
        return NULL;
    int rootnum = preorder[cur];
    cur++;
    int rootIndex = FindRootIndex(rootnum);
    //建立该根结点
    Node T = (Node)malloc(sizeof(TreeNode));
    T->num = rootnum;
    T->left = NULL;
    T->right = NULL;
    if (left != right) {
        T->left = CreateTree(left, rootIndex - 1);
        T->right = CreateTree(rootIndex + 1, right);
    }
    return T;
}


//后序遍历
int printtag = 0;
void PostTraver(Node T) {
    //int printtag = 0;
    if (T) {
        PostTraver(T->left);
        PostTraver(T->right);
        if (!printtag) {
            cout << T->num;
            printtag = 1;
        }
        else
            cout << " " << T->num;
    }
}

int main() {
    int n;
    stack<int>s;
    cin >> n;
    getchar();
    for (int i = 0; i < n*2; i++) {
        string str;
        getline(cin, str);

        if (str[1] == 'u') {
            int num = stoi(str.substr(5));
            preorder.push_back(num);
            s.push(num);
        }
        else {
            int num = s.top();
            s.pop();
            inorder.push_back(num);
        }
    }
    Node T = CreateTree(0,inorder.size()-1);
    PostTraver(T);
    return 0;
}