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));
	}
}