二叉搜索树的后序遍历序列

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。

解法是递归判断，先找根节点，划分左右子树递归求解。边界需要特殊考虑。

AC代码：

public class Solution {
public boolean VerifySquenceOfBST(int[] sequence) {
if(sequence == null || sequence.length <= 0)
return false;

int len  = sequence.length;

return solve(sequence, 0, len-1);

}

public boolean solve(int[] sequence, int start, int end) {
if(start == end)
return true;
int root = sequence[end];
//System.out.println(root);
int index = start;
for(; index < end; index ++) {
if(sequence[index] > root)
break;
}

for(int j=index; j<end; j++) {
if(sequence[j] < root)
return false;
}

boolean left = true;
if(index > start)
left = solve(sequence, start, index-1);

boolean right = true;
if(index < end)
right = solve(sequence, index, end-1);

return  (left && right) ;

}
}