【leetcode】Binary Search Tree Iterator

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.

 

最简单的，先把整棵树给遍历了，找到所有的值

 

 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  

13     vector<int> vals;

14     int index;

15     BSTIterator(TreeNode *root) {

16         DFS(root);

17         index=-1;

18     }

19    

20     void DFS(TreeNode *root)

21     {

22        

23         if(root==NULL)  return;

24  

25         DFS(root->left);

26         vals.push_back(root->val);

27         DFS(root->right);

28     }

29  

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

31     bool hasNext() {

32        

33         if(index<(int)vals.size()-1) return true;

34         else return false;

35  

36     }

37  

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

39     int next() {

40         return vals[++index];

41     }

42 };

43  

44 /**

45  * Your BSTIterator will be called like this:

46  * BSTIterator i = BSTIterator(root);

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

48  */

 

 

 

运用二叉排序树的性质

 

首先连同root把所有左边节点都push到堆栈中，

然后每当调用next时，返回堆栈栈顶元素，然后把栈顶元素的右孩子，以及他的所有左子树都压入栈即可

 

 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    

13     stack<TreeNode*> stk;

14     BSTIterator(TreeNode *root) {

15         pushLeftNode(root);

16        

17     }

18    

19     void pushLeftNode(TreeNode *root)

20     {

21  

22         while(root!=NULL)

23         {

24             stk.push(root);

25             root=root->left;

26         }

27     }

28    

29  

30  

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

32     bool hasNext() {

33        

34         return !stk.empty();

35     }

36  

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

38     int next() {

39        

40         TreeNode * smallestNode=stk.top();

41         stk.pop();

42         pushLeftNode(smallestNode->right);

43         return smallestNode->val;

44     }

45 };

46  

47 /**

48  * Your BSTIterator will be called like this:

49  * BSTIterator i = BSTIterator(root);

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

51  */