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 the BSTIterator class. The root of 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() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
int next() Moves the pointer to the right, 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() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.
- code
class BSTIterator:
# @param root, a binary search tree's root node
def __init__(self, root):
self.stack = list()
self.pushAll(root)
# @return a boolean, whether we have a next smallest number
def hasNext(self):
return self.stack
# @return an integer, the next smallest number
def next(self):
tmpNode = self.stack.pop()
self.pushAll(tmpNode.right)
return tmpNode.val
def pushAll(self, node):
while node is not None:
self.stack.append(node)
node = node.left
I use Stack to store directed left children from root. When next() be called, I just pop one element and process its right child as new root. The code is pretty straightforward. My solutions in 3 languages with Stack - LeetCode Discuss
- code
class BSTIterator:
def __init__(self, root: Optional[TreeNode]):
self.datalist = [float('-inf')]
self.inorder(root)
self.index = 0
def inorder(self, root):
if not root: return
self.inorder(root.left)
self.datalist.append(root.val)
self.inorder(root.right)
def next(self) -> int:
self.index += 1
return self.datalist[self.index]
def hasNext(self) -> bool:
if len(self.datalist) == self.index + 1:
return False
return True