bstIterator

最新推荐文章于 2022-03-01 22:00:00 发布

转载最新推荐文章于 2022-03-01 22:00:00 发布 · 447 阅读

Nice solution. It is actually breaks in-order traversal into hasNext() and next()

class BSTIterator {
public:
    stack<TreeNode *> s;
    BSTIterator(TreeNode *root)
    {
        pushAll(root);
    }
    bool hasNext()
    {
        return !s.empty();
    }
    int next()
    {
        TreeNode *tmp=s.top();
        s.pop();
        pushAll(tmp->right);
        return tem->val;
    }
    void pushAll(TreeNode*node)
    {
        while(node)
        {
            s.push(node);
            node=node->left;
        }
    }

};