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));

	}

}