LC 173. 二叉搜索树迭代器

题目点击这里

next() 和 hasNext() 操作均摊时间复杂度为 O(1) ，并使用 O(h) 内存的做法。其中 h 是树的高度，O(1)为平均时间复杂度。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
// 拆分树的非递归遍历
class BSTIterator {
public:

    stack<TreeNode*> stk;

    BSTIterator(TreeNode* root) {
        while (root) {
            stk.push(root);
            root = root->left;
        }
    }
    
    int next() {
        TreeNode* root = stk.top();
        int x = root->val;
        stk.pop();
        root = root->right;
        while (root) {
            stk.push(root);
            root = root->left;
        }
        return x;
    }
    
    bool hasNext() {
        return !stk.empty();
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */