[LeetCode] Binary Search Tree Iterator

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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

Thanks to xcv58，无需预先进行全部遍历。

利用中序遍历的递归思想：当子树的根节点访问完成后，后续节点为中序遍历该根节点右子树

 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     stack<TreeNode *> st;

13     

14     BSTIterator(TreeNode *root) {

15         pushLeft(root);

16     }

17 

18     /** @return whether we have a next smallest number */

19     bool hasNext() {

20         return !st.empty();

21     }

22 

23     /** @return the next smallest number */

24     int next() {

25         TreeNode *top = st.top();

26         st.pop();

27         pushLeft(top->right);

28         return top->val;

29     }

30     

31     void pushLeft(TreeNode *root) {

32         if (root != NULL) {

33             TreeNode *p = root;

34             st.push(p);

35             while (p->left) {

36                 st.push(p->left);

37                 p = p->left;

38             }

39         }

40     }

41 };

42 

43 /**

44  * Your BSTIterator will be called like this:

45  * BSTIterator i = BSTIterator(root);

46  * while (i.hasNext()) cout << i.next();

47  */