LeetCode //C - 173. Binary Search Tree Iterator

173. Binary Search Tree Iterator

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 [$1,10^5$].
• $0<=Node.val<=10^6$
• At most $10^5$ calls will be made to hasNext, and next.

From: LeetCode
Link: 173. Binary Search Tree Iterator

---

Solution:

Ideas：

1. During initialization (bSTIteratorCreate), perform an in-order traversal of the BST and store the nodes in a list.
2. Use an index to keep track of the current position in the list.
3. For the next method, return the node at the current index and increment the index.
4. For the hasNext method, check if the index is less than the length of the list.
5. For the free method, release any dynamically allocated memory.

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

typedef struct {
    int* nodes;
    int index;
    int size;
} BSTIterator;

void inOrder(struct TreeNode* root, int* nodes, int* index) {
    if (!root) return;
    
    inOrder(root->left, nodes, index);
    nodes[(*index)++] = root->val;
    inOrder(root->right, nodes, index);
}

BSTIterator* bSTIteratorCreate(struct TreeNode* root) {
    BSTIterator* iterator = (BSTIterator*)malloc(sizeof(BSTIterator));
    iterator->nodes = (int*)malloc(100000 * sizeof(int));  // Maximum number of nodes
    iterator->index = 0;
    iterator->size = 0;  // Initialize the size to 0

    // Fill the nodes array using inOrder traversal
    inOrder(root, iterator->nodes, &(iterator->size));

    return iterator;
}

int bSTIteratorNext(BSTIterator* obj) {
    return obj->nodes[obj->index++];
}

bool bSTIteratorHasNext(BSTIterator* obj) {
    return obj->index < obj->size;
}

void bSTIteratorFree(BSTIterator* obj) {
    free(obj->nodes);
    free(obj);
}

/**
 * Your BSTIterator struct will be instantiated and called as such:
 * BSTIterator* obj = bSTIteratorCreate(root);
 * int param_1 = bSTIteratorNext(obj);
 * bool param_2 = bSTIteratorHasNext(obj);
 * bSTIteratorFree(obj);
 */
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