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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.
Example:
BSTIterator iterator = new BSTIterator(root);
iterator.next(); // return 3
iterator.next(); // return 7
iterator.hasNext(); // return true
iterator.next(); // return 9
iterator.hasNext(); // return true
iterator.next(); // return 15
iterator.hasNext(); // return true
iterator.next(); // return 20
iterator.hasNext(); // return false
Note:next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.
难度：medium
题目：实现二叉搜索树的遍历。迭代器用根结点初始化。调用 next() 将会返回下一个最小的结点。
Runtime: 76 ms, faster than 37.23% of Java online submissions for Binary Search Tree Iterator.Memory Usage: 52.1 MB, less than 100.00% of Java online submissions for Binary Search Tree Iterator.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) {val = x;}
* }
*/
class BSTIterator {
private Stack<TreeNode> stack;
public BSTIterator(TreeNode root) {
stack = new Stack<>();
while (root != null) {
stack.push(root);
root = root.left;
}
}
/** @return the next smallest number */
public int next() {
TreeNode node = stack.pop();
TreeNode ptr = node.right;
while (ptr != null) {
stack.push(ptr);
ptr = ptr.left;
}
return node.val;
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
}
/**
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator obj = new BSTIterator(root);
* int param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/
