Description
Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):
BSTIterator(TreeNode root)Initializes an object of theBSTIteratorclass. Therootof the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.boolean hasNext()Returnstrueif there exists a number in the traversal to the right of the pointer, otherwise returnsfalse.int next()Moves the pointer to the right, then returns the number at the pointer.boolean hasPrev()Returnstrueif there exists a number in the traversal to the left of the pointer, otherwise returnsfalse.int prev()Moves the pointer to the left, then returns the number at the pointer.
Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.
You may assume that next() and prev() calls will always be valid. That is, there will be at least a next/previous number in the in-order traversal when next()/prev() is called.
Example 1:
Input ["BSTIterator", "next", "next", "prev", "next", "hasNext", "next", "next", "next", "hasNext", "hasPrev", "prev", "prev"] [[[7, 3, 15, null, null, 9, 20]], [null], [null], [null], [null], [null], [null], [null], [null], [null], [null], [null], [null]] Output [null, 3, 7, 3, 7, true, 9, 15, 20, false, true, 15, 9] Explanation // The underlined element is where the pointer currently is. BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); // state is [3, 7, 9, 15, 20] bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 3 bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 7 bSTIterator.prev(); // state becomes [3, 7, 9, 15, 20], return 3 bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 7 bSTIterator.hasNext(); // return true bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 9 bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 15 bSTIterator.next(); // state becomes [3, 7, 9, 15, 20], return 20 bSTIterator.hasNext(); // return false bSTIterator.hasPrev(); // return true bSTIterator.prev(); // state becomes [3, 7, 9, 15, 20], return 15 bSTIterator.prev(); // state becomes [3, 7, 9, 15, 20], return 9
Constraints:
- The number of nodes in the tree is in the range
[1, 105]. 0 <= Node.val <= 106- At most
105calls will be made tohasNext,next,hasPrev, andprev.
Follow up: Could you solve the problem without precalculating the values of the tree?
Solutions
Solution 1: In-order Traversal + Array
We can use in-order traversal to store the values of all nodes in the binary search tree into an array nums, and then use the array to implement the iterator. We define a pointer i, initially i = -1, which points to an element in the array nums. Each time we call next(), we add 1 to the value of i and return nums[i]; each time we call prev(), we subtract 1 from the value of i and return nums[i].
In terms of time complexity, initializing the iterator requires O(n) time, where n is the number of nodes in the binary search tree. Each call to next() and prev() requires O(1) time. In terms of space complexity, we need O(n) space to store the values of all nodes in the binary search tree.
# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class BSTIterator: def __init__(self, root: Optional[TreeNode]): self.nums = [] def dfs(root): if root is None: return dfs(root.left) self.nums.append(root.val) dfs(root.right) dfs(root) self.i = -1 def hasNext(self) -> bool: return self.i < len(self.nums) - 1 def next(self) -> int: self.i += 1 return self.nums[self.i] def hasPrev(self) -> bool: return self.i > 0 def prev(self) -> int: self.i -= 1 return self.nums[self.i] # Your BSTIterator object will be instantiated and called as such: # obj = BSTIterator(root) # param_1 = obj.hasNext() # param_2 = obj.next() # param_3 = obj.hasPrev() # param_4 = obj.prev()(code-box)
