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

算法分析：首先假设输入的二叉树是BST，并且不是镜像的BST。利用先序遍历找出根节点，然后递归其左子树和右子树，接着就得出其后续遍历。如果其后序遍历的长度不足n，那么再假设输入的二叉树是镜像的BST，再得出后序遍历，如果其后序遍历的长度等于n，那么就输出YES，否则输出NO。

#include <stdio.h>#include <stdlib.h>#include <string.h>int pre[1000], post[1000];int flag = 1, lenpost = 0;void solve(int root, int tail) {    if (root > tail) return;    int i = root + 1, j = tail;    if (flag)     { //正常的BST        for (; i <= tail && pre[root] > pre[i]; i++);        for (; j > root && pre[root] <= pre[j]; j--);    }    else     { //镜像的BST        for (; i <= tail && pre[root] <= pre[i]; i++);        for (; j > root && pre[root] > pre[j]; j--);    }    if (i - j != 1) return;    solve(root + 1, j); //类似后序遍历    solve(i, tail);    post[lenpost++] = pre[root];}int main(){    int n, i;    scanf("%d", &n);    for (i = 0; i < n; i++)        scanf("%d", &pre[i]);    solve(0, n - 1);    if (lenpost != n)     {        flag = 0;        for (i = 0; i < n; i++)            post[i] = 0;        lenpost = 0;        solve(0, n - 1);    }    if (lenpost == n)     {        printf("YES\n%d", post[0]);        for (i = 1; i < n; i++)            printf(" %d", post[i]);    }    else {        printf("NO");    }    system("pause");    return 0;}