Binary Search Tree Iterator
描述
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.
分析
考察非递归的中序遍历。这道题本质上是写一个二叉树的中序遍历的迭代器。内部设置一个栈,初始化的时候,存储从根节点到最左叶子节点的路径。在遍历的过程中,每次从栈中弹出一个元素,作为当前的返回结果,同时探测一下当前节点是否存在右孩子,如果有,则进入右孩子,并把从该右孩子到最左叶子节点的所有节点入栈。
代码
// Binary Search Tree Iterator
public class BSTIterator {
public BSTIterator(TreeNode root) {
stack = new Stack<>();
while (root != null) {
stack.push(root);
root = root.left;
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
final TreeNode node = stack.pop();
if (node.right != null) {
TreeNode p = node.right;
while (p != null) {
stack.push(p);
p = p.left;
}
}
return node.val;
}
private Stack<TreeNode> stack;
}