PAT 1086. Tree Traversals Again (25)

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

 

#include <iostream>
#include <string>
#include <stack>

using namespace std;

struct Node
{
    int value;
    int left = 0;
    int right = 0;
};

Node tree[31];    //we use tree[0] to devote the null pointer
int treeroot;

void PostTraversal(int root)
{
    if (tree[root].value == 0)
        return;
    PostTraversal(tree[root].left);
    PostTraversal(tree[root].right);
    cout << tree[root].value;
    if (tree[root].value != treeroot)
        cout << " ";
}

int main()
{
    int n;
    cin >> n;
    stack<int> s;

    string op, lastOp;
    int value, lastValue, root;
    cin >> op >> value;    //fist operation must be push
    //construct the tree
    treeroot = root = value;
    tree[value].value = value;
    s.push(value); 
    lastOp = "Push";
    lastValue = value;
    n *= 2;
    while (--n)
    {
        cin >> op;
        if (op == "Push")    //push
        {
            cin >> value;
            tree[value].value = value;
            s.push(value);
            if (lastOp == "Push")
                tree[lastValue].left = value;
            else
                tree[lastValue].right = value;
            lastValue = value;
            lastOp = "Push";
        }
        else    //pop
        {
            lastValue = s.top();
            s.pop();    
            lastOp = "Pop";
        }
    }

    PostTraversal(root); 
}