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.
Analyse: Inorder traversal of BST.
Runtime: 28ms.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class BSTIterator { 11 public: 12 BSTIterator(TreeNode *root) { 13 inorder(root); 14 } 15 16 /** @return whether we have a next smallest number */ 17 bool hasNext() { 18 return !qu.empty(); 19 } 20 21 /** @return the next smallest number */ 22 int next() { 23 int nextSmallest = qu.front(); 24 qu.pop(); 25 return nextSmallest; 26 } 27 private: 28 queue<int> qu; 29 30 void inorder(TreeNode* root) { 31 if(!root) return; 32 33 inorder(root->left); 34 qu.push(root->val); 35 inorder(root->right); 36 } 37 }; 38 39 /** 40 * Your BSTIterator will be called like this: 41 * BSTIterator i = BSTIterator(root); 42 * while (i.hasNext()) cout << i.next(); 43 */