PAT-A-1086. Tree Traversals Again (25)

1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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<stack>
#include<cstdio>
#include<string.h>

using namespace std;

const int maxn = 50;
struct node{
int date;
node *lchild;
node *rchild;
};

int pre[maxn];
int in[maxn];
int n;

node *create(int prel, int prer, int inl, int inr)
{

if (prel > prer)
return NULL;

node *root = new node;
root->date = pre[prel];

int k;
for (k = inl; k <= inr; k++)
{
if (in[k] == pre[prel])
break;
}
int len = k - inl;
root->lchild = create(prel + 1, prel + len, inl, k - 1);
root->rchild = create(prel + len + 1, prer, k + 1, inr);

return root;

}

int num = 0;
void postorder(node *root)
{
if (root == NULL)
return;

postorder(root->lchild);
postorder(root->rchild);
cout << root->date;

num++;
if (num < n)
{
cout << " ";
}

}

int main()
{
cin >> n;
char s[5];
stack<int> st;
int num1 = 0;
int num2 = 0;
int d;
for (int i = 0; i < 2 * n; i++)
{
cin >> s;
if (strcmp(s, "Push")==0)
{
cin >> d;
st.push(d);
pre[num1++] = d;
}
else
{
in[num2++] = st.top();
st.pop();
}

}

node *root = create(0, n - 1, 0, n - 1);
postorder(root);

system("pause");
return 0;
}