二叉搜索树的后续遍历序列

/**
* 输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历结果，假设输入的数组的任意两个数字都互不相同
* @author Q.Yuan
*
*/
public class JudgePostOrder {
/**
*
* @param a    后序遍历结果
* @param low  序列的开始
* @param high 序列的结束
* @return
*/
public boolean isPostOrder(int[] a,int low,int high){
int root = a[high];
int leftEnd = low;
//找到第一个比根节点大的值，其为右子树序列的开始
for(;leftEnd <= high - 1;leftEnd++){
if(a[leftEnd] > root){
break;
}
}
leftEnd--;
//判断左子树的值是否都小于根节点的值
for(int i = low;i <= leftEnd;i++ ){
if(a[i] > root){
return false;
}
}
//判断右子树的值是否都大于根节点的值
for(int j = leftEnd + 1;j <= high - 1;j++){
if(a[j] < root){
return false;
}
}
boolean left = true ;
boolean right = true;
//如果有左子树
if(low <= leftEnd){
left = isPostOrder(a, low, leftEnd);
}
//如果有右子树
if(leftEnd < high - 1){
right = isPostOrder(a, leftEnd + 1, high - 1);
}
return left && right;
}
public static void main(String[] args) {
JudgePostOrder jpo = new JudgePostOrder();
//		int[] a = {5,7,6,9,11,10,8};
//		int[] a = {1,2,3,8,7,6,5};
int[] a = {1,2,3,5};
System.out.println(jpo.isPostOrder(a, 0, a.length - 1));
}
}