剑指offer——二叉搜索树的后序遍历序列

 1 public class Solution {
 2     public boolean VerifySquenceOfBST(int [] sequence) {
 3         if (sequence.length == 0)//空树
 4             return false;
 5         if (sequence.length == 1)//单个节点树
 6             return true;
 7         return judge(sequence, 0, sequence.length-1);
 8     }
 9     public boolean judge(int[] a, int star, int end){
10         if(star>=end){//迭代终止
11             return true;
12         }
13         int i = star;
14         while(a[i]<a[end]){//左子树
15             i++;
16         }
17         for(int j = i;j<end;j++){//右子树
18             if(a[j]<a[end])
19                 return false;
20         }
21         return(judge(a, star, i-1)&&judge(a, i,end-1));//对左子树和右子树进行判断
22     }
23 }