173. Binary Search Tree Iterator - Medium

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.

 

M1: 先inorder全部遍历完（慢）

time: O(n) , space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class BSTIterator {
    List<Integer> list;
    int idx;

    public BSTIterator(TreeNode root) {
        list = new ArrayList<>();
        idx = 0;
        inorder(root, list);
    }
    
    public void inorder(TreeNode root, List<Integer> list) {
        if(root == null) {
            return;
        }
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }
    
    /** @return the next smallest number */
    public int next() {
        int tmp = idx;
        idx++;
        return list.get(tmp);
    }
    
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return idx < list.size();
    }
}

/**
 * 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();
 */

 

M2: optimized，用stack，先把所有的左边的节点压进栈，当call next()时再按顺序出栈

next()　　　   -- time: O(h), space: O(h)

hasNext()　　-- time: O(1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class BSTIterator {
    Stack<TreeNode> stack = new Stack<>();

    public BSTIterator(TreeNode root) {
        getAllLeft(root);
    }
    
    public void getAllLeft(TreeNode root) {
        while(root != null) {
            stack.push(root);
            root = root.left;
        }
    }
    
    /** @return the next smallest number */
    public int next() {
        TreeNode node = stack.pop();
        getAllLeft(node.right);       
        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();
 */