Binary Search Tree Iterator

Description

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

• BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
• boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
• int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

Example 1:

Input [“BSTIterator”, “next”, “next”, “hasNext”, “next”, “hasNext”, “next”, “hasNext”, “next”, “hasNext”] [7, 3, 15, null, null, 9, 20, [], [], [], [], [], [], [], [], []] Output [null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); bSTIterator.next(); // return 3 bSTIterator.next(); // return 7 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 9 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 15 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 20 bSTIterator.hasNext(); // return False

Constraints:

• The number of nodes in the tree is in the range [1, 105].
• 0 <= Node.val <= 106
• At most 105 calls will be made to hasNext, and next.

Follow up:

• Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?

Code

$O(n)$ space

在 constructer 中使用 Binary Tree Inorder Traversal 中的方法，建立 inorder traversal vector，後續的 next 移動就是 vector 的 index 存取問題。

/**
 * 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) {}
 * };
 */
class BSTIterator {
    vector<int> inorder;
    int idx;
public:
    BSTIterator(TreeNode* root) {
        idx = -1;
        stack<TreeNode*> st;
        while(root || !st.empty()) {
            while(root) {
                st.push(root);
                root = root->left;
            }
            auto node = st.top();
            st.pop();
            inorder.push_back(node->val);
            root = node->right;
        }
    }
    
    int next() {
        idx++;
        return inorder[idx];
    }
    
    bool hasNext() {
        return idx < 0 || idx < inorder.size() - 1;
    }
};
 
/**
 * 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();
 */

$O(h)$ space, h is the tree height

將上述的 inorder traversal 拆解成 partial inorder traversal，就可以使 space complexity 降到 $O(h)$。

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

Source

• Binary Search Tree Iterator - LeetCode