1086 Tree Traversals Again

Posted 2021-03-02 ruruozhenhao

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

Code：

#include<iostream>
#include<queue>
#include<stack>

using namespace std;

typedef struct Node *node;
struct Node {
    int value;
    node leftSon;
    node rightSon;
    node father;
    Node():value(), leftSon(), rightSon(), father(){}
};

void postOrder(node h) {
    if (h != NULL) {
        postOrder(h->leftSon);
        postOrder(h->rightSon);
        cout << h->value << " ";
    }
}

int main() {
    int n;
    cin >> n;
    getchar();
    string str, op, num;
    int pos;
    queue<string> q;
    stack<int> s;
    stack<string> leftOrRight;
    for (int i = 0; i < n*2; ++i) {
        getline(cin, str);
        if (str[1] == ‘u‘) {
            pos = str.find(‘ ‘);
            op = str.substr(0, pos);
            num = str.substr(pos+1);
            q.push(op);
            q.push(num);
        } else {
            q.push(str);
        }
    }

    node head = new Node();
    node prt = head;
    while (!q.empty()) {
        if (q.front() == "Push") {
            q.pop();
            int value = stoi(q.front());
            s.push(value);
            q.pop();
            node temp = new Node();
            if (prt->leftSon) {
                leftOrRight.push("right");
                prt->rightSon = temp;
                temp->father = prt;
            } else {
                leftOrRight.push("left");
                prt->leftSon = temp;
                temp->father = prt;         
            }
            prt = temp;
        } else {
            q.pop();
            if (leftOrRight.top() == "left") prt = prt->father;
            prt->value = s.top();
            if (leftOrRight.top() == "right") prt = prt->father;
            s.pop();
            leftOrRight.pop();
        }
    }
    postOrder(prt);
    return 0;
}