面试题24：二叉搜索树后序遍历

判断左子树小于根节点，右子树大于根节点，然后递归

bool VerifySquenceOfBST(int sequence[], int length)
{
    if(sequence == NULL || length <= 0)
        return false;

    int root = sequence[length - 1];

    // 在二叉搜索树中左子树的结点小于根结点
    int i = 0;
    for(; i < length - 1; ++ i)
    {
        if(sequence[i] > root)
            break;
    }

    // 在二叉搜索树中右子树的结点大于根结点
    int j = i;
    for(; j < length - 1; ++ j)
    {
        if(sequence[j] < root)
            return false;
    }

    // 判断左子树是不是二叉搜索树
    bool left = true;
    if(i > 0)
        left = VerifySquenceOfBST(sequence, i);

    // 判断右子树是不是二叉搜索树
    bool right = true;
    if(i < length - 1)
        right = VerifySquenceOfBST(sequence + i, length - i - 1);

    return (left && right);
}

 用容器

bool VerifySquenceOfBST(vector<int> sequence) {
    int len = sequence.size();
    if (len <= 0)
        return false;
    if (len <= 2)
        return true;
    int root = sequence[len - 1];
    int i = 0;
    for (; i < len - 1; i++)
    {
        if (sequence[i] > root)
            break;
    }
    int j = i;    //分割
    for (; j < len - 1; j++)
    {
        if (sequence[j] < root)
            return false;
    }
    bool left = true;
    if (i > 0)
    {
        vector<int> sequence_left(sequence.begin(), sequence.begin() + i);
        left = VerifySquenceOfBST(sequence_left);
    }

    bool right = true;
    if (i < len - 1)
    {
        vector<int> sequence_right(sequence.begin() + i, sequence.end() - 1);
        right = VerifySquenceOfBST(sequence_right);
    }
    return left && right;
}