Binary Search Tree Iterator

Desicription

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Solution

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/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/
class BSTIterator {
private:
stack<TreeNode* > nodeStack;
void findLeft(TreeNode* cur) {
while(cur)
nodeStack.push(cur), cur = cur->left;
}
public:
BSTIterator(TreeNode *root) {
findLeft(root);
}
/** @return whether we have a next smallest number */
bool hasNext() {
return nodeStack.size();
}
/** @return the next smallest number */
int next() {
TreeNode* cur = nodeStack.top();
nodeStack.pop();
if(cur->right)
findLeft(cur->right);
return cur->val;
}
};