"Leetcode" Binary Search Tree Iterator (2 solutions)

Binary Search Tree Iterator

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

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

Note: next() hasNext() and 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.

Solution One:

The sequential output is loaded in the queue after the sequence traversal.

/** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * T Reenode (int x): Val (x), left (null), right (NULL) {} *}; */classBstiterator { Public: Queue<int>minq; Map<treenode*,BOOL>m; Stack<treenode *>s; Bstiterator (TreeNode*root) {        //inorder Traversal        if(Root! =NULL)            {S.push (root); M[root]=true;  while(!S.empty ()) {TreeNode* top =S.top (); if(Top->left && m.find (top->left) = =M.end ()) {S.push (top-Left ); M[top->left] =true; Continue; } minq.push (Top-val);                S.pop (); if(Top->right && m.find (top->right) = =M.end ()) {S.push (top-Right ); M[top->right] =true; }            }        }    }    /** @return Whether we have a next smallest number*/    BOOLHasnext () {return!Minq.empty (); }    /** @return The next smallest number*/    intNext () {intFront =Minq.front ();        Minq.pop (); returnFront; }};/** * Your bstiterator'll be called like this: * bstiterator i = bstiterator (root), * while (I.hasnext ()) cout << ; I.next (); */

Solution Two:

Thanks to xcv58, there is no need to advance all the traversal.

Recursive thinking using the middle sequence traversal: When the root node of the subtree is accessed, the subsequent nodes are traversing the right subtree of the root node for the middle sequence.

/** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * T Reenode (int x): Val (x), left (null), right (NULL) {} *}; */classBstiterator { Public: Stack<treenode *>s; Bstiterator (TreeNode*root)    {pushleft (root); }    /** @return Whether we have a next smallest number*/    BOOLHasnext () {return!S.empty (); }    /** @return The next smallest number*/    intNext () {TreeNode* top =S.top ();        S.pop (); Pushleft (Top-Right ); returnTop->Val; }        voidPushleft (treenode*root) {        if(Root! =NULL)            {S.push (root); TreeNode* cur =Root;  while(cur->Left ) {S.push (cur-Left ); Cur= cur->Left ; }        }    }};/** * Your bstiterator'll be called like this: * bstiterator i = bstiterator (root), * while (I.hasnext ()) cout << ; I.next (); */

"Leetcode" Binary Search Tree Iterator (2 solutions)