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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.
Example 1:
Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]
Explanation
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // return 3
bSTIterator.next(); // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 20
bSTIterator.hasNext(); // return False
Constraints:
[1, 105]
.0 <= Node.val <= 106
105
calls will be made to hasNext
, and next
.Follow up:
next()
and hasNext()
to run in average O(1)
time and use O(h)
memory, where h
is the height of the tree?public class BSTIterator {
private Stack<TreeNode> stack = new Stack<TreeNode>();
public BSTIterator(TreeNode root) {
pushAll(root);
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
TreeNode tmpNode = stack.pop();
pushAll(tmpNode.right);
return tmpNode.val;
}
private void pushAll(TreeNode node) {
for (; node != null; stack.push(node), node = node.left);
}
}
class BSTIterator {
stack<TreeNode *> myStack;
public:
BSTIterator(TreeNode *root) {
pushAll(root);
}
/** @return whether we have a next smallest number */
bool hasNext() {
return !myStack.empty();
}
/** @return the next smallest number */
int next() {
TreeNode *tmpNode = myStack.top();
myStack.pop();
pushAll(tmpNode->right);
return tmpNode->val;
}
private:
void pushAll(TreeNode *node) {
for (; node != NULL; myStack.push(node), node = node->left);
}
};
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
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