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.
Approach #1: C++.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 | / * * * Definition for binary tree * struct TreeNode { * int val; * TreeNode * left; * TreeNode * right; * TreeNode( int x) : val(x), left(NULL), right(NULL) {} * }; * / class BSTIterator { public: BSTIterator(TreeNode * root) { helper(root); } / * * @ return whether we have a next smallest number * / bool hasNext() { if (minNums.empty()) return false; else return true; } / * * @ return the next smallest number * / int next () { TreeNode * cur = minNums.top(); minNums.pop(); helper(cur - >right); return cur - >val; } private: stack<TreeNode * > minNums; void helper(TreeNode * root) { while (root ! = nullptr) { minNums.push(root); root = root - >left; } } }; / * * * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i. next (); * / |
Approach #2: Java.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 | / * * * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode( int x) { val = x; } * } * / public class BSTIterator { public BSTIterator(TreeNode root) { helper(root); } / * * @ return whether we have a next smallest number * / public boolean hasNext() { if (stack.isEmpty()) return false; return true; } / * * @ return the next smallest number * / public int next () { TreeNode cur = stack.pop(); helper(cur.right); return cur.val; } private Stack<TreeNode> stack = new Stack<TreeNode>(); private void helper(TreeNode root) { while (root ! = null) { stack.push(root); root = root.left; } } } / * * * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.hasNext()) v[f()] = i. next (); * / |
Approach #3: Python.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 | # Definition for a binary tree node # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class BSTIterator( object ): def __init__( self , root): """ :type root: TreeNode """ self .stack = list () self .helper(root) def hasNext( self ): """ :rtype: bool """ return self .stack def next ( self ): """ :rtype: int """ cur = self .stack.pop() self .helper(cur.right) return cur.val def helper( self , root): while root is not None : self .stack.append(root) root = root.left # Your BSTIterator will be called like this: # i, v = BSTIterator(root), [] # while i.hasNext(): v.append(i.next()) |
永远渴望,大智若愚(stay hungry, stay foolish)
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】博客园社区专享云产品让利特惠,阿里云新客6.5折上折
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步