剑指Offer22 判断数组是否为某二叉搜索树的后序遍历

 /*************************************************************************
     > File Name: 22_SequenceOfBST.cpp
     > Author: Juntaran
     > Mail: JuntaranMail@gmail.com
     > Created Time: 2016年08月30日 星期二 20时34分33秒
  ************************************************************************/

 #include <stdio.h>
 #include <stdlib.h>

 // 判断一个数组是不是某个二叉搜索树的后序遍历
 bool isSequenceOfBST(int* sequence, int length)
 {
     if (sequence==NULL || length<=)
         return false;

     int root = sequence[length-];

     int i, j;

     // 二叉搜索树的左子树都小于根
     for (i = ; i < length-; ++i)
     {
         if (sequence[i] > root)
             break;
     }

     // 二叉搜索树的右子树都大于根
     for (j = i; j < length-; ++j)
     {
         if (sequence[j] < root)
             return false;
     }

     // 递归判断左子树是不是二叉搜索树
     bool left = true;
     if (i > )
         left = isSequenceOfBST(sequence, i);

     // 递归判断右子树是不是二叉搜索树
     bool right = true;
     if (i < length-)
         right = isSequenceOfBST(sequence+i, length-i-);

     return (left && right);
 }

 int main()
 {
     int sequence1[] = {,,,,,,};
     int sequence2[] = {,,,};
     bool ret1 = isSequenceOfBST(sequence1, );
     bool ret2 = isSequenceOfBST(sequence2, );
     if (ret1 == true)
         printf("ret1 is true\n");
     else
         printf("ret1 is false\n");

     if (ret2 == true)
         printf("ret2 is true\n");
     else
         printf("ret2 is false\n");

     return ;
 }