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 = ;

struct Node

{

    int data;

    Node *lchild, *rchild;

};

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

int num = ;

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 = , inIndex = ;

    char str[];

    stack<int> s;

    for (int i = ; i < *n; i++)

    {

        getchar();

        scanf("%s",str);

        if (  == strcmp(str,"Push"))

        {

            scanf("%d",&x);

            s.push(x);

            pre[preIndex++] = x;

        }

        else

        {

            x = s.top();

            s.pop();

            in[inIndex++] = x;

        }

    }

    Node *root = createTree(,n-,,n-);

    postOrder(root,n);

    return ;

}

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+, preL+numLeft, inL, k-);

    root->rchild = createTree(preL+numLeft+, preR, k+, 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(" ");

    }

}

巴特西

03-树3 Tree Traversals Again (25 分)

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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