Problem #
Source: [PAT 1086]
Description #
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
- can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
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
PopSample Output: #
3 4 2 6 5 1Solution #
Code #
#include <string>
#include <iostream>
#include <stack>
#define max_size 31
using namespace std;
int n, pre[max_size], in[max_size];
struct node
{
int v;
node *left;
node*right;
};
void init()
{
string item;
cin >> n;
stack<int> s;
int a = 0, b = 0, num;
for (int i = 0; i < 2 * n; i++)
{
cin >> item;
if (item[1] == 'o') //pop
{
in[b++] = s.top();
s.pop();
}
else // push
{
cin>>num;
pre[a++] = num;
s.push(num);
}
}
}
node *make(int inleft, int inright, int preleft, int preright)
{
if (preleft > preright)
return NULL;
node *root = new node;
//root->left =root->right= NULL;
root->v = pre[preleft];
int i = inleft;
for ( i = inleft; i <= inright; i++)
{
if (in[i] == pre[preleft])
break;
}
root->left = make(inleft, i - 1, preleft+1, preleft + i - inleft );
root->right = make(i + 1, inright, preleft + i - inleft+1, preright);
return root;
}
void postOrder(node *root)
{
if (root == NULL)
return;
postOrder(root->left);
postOrder(root->right);
cout << root->v;
if (n > 1)
cout << " ";
--n;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
init();
node *root = make(0, n - 1, 0, n - 1);
postOrder(root);
return 0;
}
