03-树3 Tree Traversals Again (25 分)

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

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
const int maxn = 50;

struct Node
{
    int data;
    Node *lchild, *rchild;
};

int in[maxn] = {0}, pre[maxn] = {0};
int num = 0;

Node*  createTree(int preL, int preR, int inL, int inR);
void postOrder(Node *root,int n);

int main()
{
    int n;
    scanf("%d",&n);
    
    int x;
    int preIndex = 0, inIndex = 0;
    char str[5];
    stack<int> s;
    
    for (int i = 0; i < 2*n; i++)
    {
        getchar();
        scanf("%s",str);
        if ( 0 == strcmp(str,"Push"))
        {
            scanf("%d",&x);
            s.push(x);
            pre[preIndex++] = x;
        }
        else
        {
            x = s.top();
            s.pop();
            in[inIndex++] = x;
        }
    }
    
    Node *root = createTree(0,n-1,0,n-1);
    postOrder(root,n);
    return 0;
}

Node*  createTree(int preL, int preR, int inL, int inR)
{
    if (preL > preR)
    {
        return NULL;
    }
    
    Node *root = new Node;
    root->data = pre[preL];
    
    int k;
    for (k = inL; k <= inR; k++)
    {
        if (in[k] == pre[preL])
        {
            break;
        }
    }
    
    int numLeft = k - inL;
    root->lchild = createTree(preL+1, preL+numLeft, inL, k-1);
    root->rchild = createTree(preL+numLeft+1, preR, k+1, inR);
    return root;
}

void postOrder(Node *root,int n)
{
    if (root == NULL)
    {
        return;
    }
    postOrder(root->lchild,n);
    postOrder(root->rchild,n);
    printf("%d",root->data);
    
    num++;
    if (num < n)
    {
        printf(" ");
    }
}