剑指offer--二叉搜索树的后序遍历序列

题目描述
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的结果。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同

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思路：
二叉搜索树的性质：根节点大于左子树所有元素，小于右子数的所有元素。后序遍历的话，最后一个元素就为根节点root。若数组为null或者长度为0，则返回true，若数组基本有序（即所有元素已经是升序或降序），返回true，否则，找到划分左右子树的数组元素下标（sequence[i] < root && sequence[i+1] > root），之后判断下标属于[0,i]的子序列是否全小于root，[i,len-2]的子序列是否全大于root。

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AC代码：

import java.util.Arrays;public class Solution {    public boolean VerifyLeftTree(int[] leftsequence , int root){        boolean flag = true;        for ( int i = 0 ; i < leftsequence.length ; i++){            if ( leftsequence[i] >= root){                flag = false;                break;            }        }        return flag;    }    public boolean VerifyRightTree(int[] rightsequence , int root){        boolean flag = true;        for ( int i = 0 ; i < rightsequence.length ; i++){            if ( rightsequence[i] <= root){                flag = false;                break;            }        }        return flag;    }    //判断sequence是否已经有序    public boolean IsOrder(int [] sequence){        //判断是否已经是升序        int[] tmp = Arrays.copyOfRange(sequence, 0, sequence.length);        Arrays.sort(tmp);        boolean flag1 = true;        for(int i = 0 ; i < sequence.length ; i++){            if ( sequence[i] != tmp[i]){                flag1 = false;                break;            }        }        //判断是否已经是降序        int cnt = 0;        boolean flag2 = true;        for ( int i = tmp.length-1 ; i >= 0 ; i--){            if ( sequence[cnt++] != tmp[i]){                flag2 = false;                break;            }        }        boolean flag = flag1 || flag2;        return flag;    }    public boolean VerifySquenceOfBST(int [] sequence) {        int partition = 0;        int len = sequence.length;        if ( sequence == null || len == 0){            return false;        }        int root = sequence[len-1];        if ( IsOrder(sequence)){            return true;        }        for ( int i = 0 ; i < len-1 ; i++){            if (sequence[i] < root && sequence[i+1] > root){                partition = i;                break;            }        }        int[] left = Arrays.copyOfRange(sequence, 0, partition+1);        int[] right = Arrays.copyOfRange(sequence, partition+1, len-1);        for ( int i = 0 ; i < left.length ; i++){            System.out.print(left[i]+" ");        }        System.out.println();        for ( int i = 0 ; i < right.length ; i++){            System.out.print(right[i]+" ");        }        boolean leftflag = VerifyLeftTree(left,root);        boolean rightflag = VerifyRightTree(right,root);        boolean flag = leftflag & rightflag;        return flag;    }}