PAT A1043 Is It a Binary Search Tree

PAT A1043 Is It a Binary Search Tree(25分)

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

If we swap the left and right subtrees of every node, then the resulting tree is called theMirror Imageof a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integerN(≤1000). ThenNinteger keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a lineYESif the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, orNOif not. Then if the answer isYES, print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

7
8 6 5 7 10 8 11

Sample Output 1:

YES
5 7 6 8 11 10 8

Sample Input 2:

7
8 10 11 8 6 7 5

Sample Output 2:

YES
11 8 10 7 5 6 8

Sample Input 3:

7
8 6 8 5 10 9 11

Sample Output 3:

NO

注意點

• 本題的意思是將輸入的序列構造爲二叉查找樹，並判斷該序列是否與該樹的先序遍歷或先序遍歷的鏡像一致，一致則輸出yes，並打印該樹的後序遍歷或後序遍歷的鏡像。
• 起初比較兩個序列是否相同時，考慮的是通常數組，但這種方法比較麻煩。使用vector型數組能夠直接比較，方便使用。

代碼

#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;

struct node
{
    int data;
    node* lchild;
    node* rchild;
};

node* newNode(int x)
{
    node* root = new node;
    root -> data = x;
    root -> lchild = NULL;
    root -> rchild = NULL;
    return root;
}

void insert(node* &root , int x)
{
    if (root == NULL)
    {
        root = newNode(x);
        return;
    }
    if (x < root -> data)
        insert(root -> lchild, x);
    else
        insert(root -> rchild, x);
}

//先序 
void preOrder(node* root, vector<int>& vi)
{
    if (root == NULL) return;
    vi.push_back(root -> data);
    preOrder(root -> lchild, vi);
    preOrder(root -> rchild, vi);
}

//鏡像先序
void preOrderMirror(node* root, vector<int>& vi) 
{
    if (root == NULL) return;
    vi.push_back(root -> data);
    preOrderMirror(root -> rchild, vi);
    preOrderMirror(root -> lchild, vi);
}
//後續遍歷
void postOrder(node* root, vector<int>& vi) 
{
    if (root == NULL)    return;
    postOrder(root -> lchild, vi);
    postOrder(root -> rchild, vi);
    vi.push_back(root -> data);
}
// 鏡像樹後序 
void postOrderMirror(node* root, vector<int>& vi) 
{
    if (root == NULL) return;
    postOrderMirror(root -> rchild, vi);
    postOrderMirror(root -> lchild, vi);
    vi.push_back(root -> data);
}

vector<int> origin, pre, preM, post, postM;

int main()
{
    int n, data;
    node* root = NULL;
    scanf("%d", &n);
    for (int i = 0; i < n; i ++)    
    {
        scanf("%d", &data);
        origin.push_back(data);
        insert(root, data);
    }
    preOrder(root, pre);
    preOrderMirror(root, preM);
    
    postOrder(root, post);
    postOrderMirror(root, postM);
    if (origin == pre)
    {
        printf("YES\n");
        for (int i = 0; i < post.size(); i ++)
        {
            printf("%d", post[i]);
            if ( i < post.size() - 1)
                printf(" ");
        }
    }
    else if (origin == preM)
    {
        printf("YES\n");
        for (int i = 0; i < postM.size(); i ++)
        {
            printf("%d", postM[i]);
            if (i < postM.size() - 1)
                printf(" ");
        }
    }
    else
    {
        printf("NO\n");
    }
    return 0;
}