1043. Is It a Binary Search Tree (25)

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 the Mirror Image of 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 integer N (<=1000). Then N integer 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 line “YES” if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or “NO” if not. Then if the answer is “YES”, 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

题目大意：按照题目给出的序列，生成一颗二叉搜索树，然后判断给出的序列是否为树的前序或者镜像，是的话输出后序。
思路：生成树，然后判断就可以了。

#include <iostream>using namespace std;const int maxn = 1005;typedef struct Node *Position;struct Node{    int x;    Position left;    Position right;};int n, cnt, node[maxn], pre[maxn], mirror[maxn], post[maxn];Position insert(Position T, int x){    if(T == NULL){        T = new Node();        if(T == NULL)            printf("Out of space!!!");        else {            T->x = x;            T->left = NULL;            T->right = NULL;        }    } else if(x < T->x)        T->left = insert(T->left, x);    else if(T->x <= x)        T->right = insert(T->right, x);    return T;}void preOrder(Position root){    if(root == NULL)            return ;    else {        pre[cnt++] = root->x;        preOrder(root->left);        preOrder(root->right);    }}void mirrorOrder(Position root){    if(root == NULL)        return ;    else {        mirror[cnt++] = root->x;        mirrorOrder(root->right);        mirrorOrder(root->left);    }}bool isPre(){    for(int i = 0; i < n; ++i){        if(pre[i] != node[i])            return false;    }    return true;}bool isMirror(){    for(int i = 0; i < n; ++i)        if(mirror[i] != node[i])            return false;    return true;}void postOrder(Position root){    if(root == NULL)        return ;    else {        postOrder(root->left);        postOrder(root->right);        post[cnt++] = root->x;    }}void mpostOrder(Position root){    if(root == NULL)        return ;    else {        mpostOrder(root->right);        mpostOrder(root->left);        post[cnt++] = root->x;    }}void printPost(){    for(int i = 0; i < n; ++i){        if(i)            printf(" ");        printf("%d", post[i]);    }}int main(){    int x;    Position Bst = NULL;    scanf("%d", &n);    for(int i = 0; i < n; ++i){        scanf("%d", &x);            node[i] = x;        Bst = insert(Bst, x);       }    preOrder(Bst);    if(isPre()){        printf("YES\n");        cnt = 0;        postOrder(Bst);        printPost();    } else {        cnt = 0;        mirrorOrder(Bst);        if(isMirror()){            printf("YES\n");            cnt = 0;            mpostOrder(Bst);            printPost();        } else             printf("NO\n");    } }