Leetcode 173 二叉搜索树迭代器

题目定义：

实现一个二叉搜索树迭代器。你将使用二叉搜索树的根节点初始化迭代器。

调用 next() 将返回二叉搜索树中的下一个最小的数。

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // 返回 3
iterator.next();    // 返回 7
iterator.hasNext(); // 返回 true
iterator.next();    // 返回 9
iterator.hasNext(); // 返回 true
iterator.next();    // 返回 15
iterator.hasNext(); // 返回 true
iterator.next();    // 返回 20
iterator.hasNext(); // 返回 false

方式一(运用栈)：

class BSTIterator {
    private Deque<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new ArrayDeque<>();
        TreeNode cur = root;
        while(cur != null){
            if(cur.left == null){
                stack.add(new TreeNode(cur.val));
                cur = cur.right;
            }else{
                TreeNode pre = cur.left;
                while(pre.right != null && pre.right != cur){
                    pre = pre.right;
                }
                if(pre.right == null){
                    pre.right = cur;
                    cur = cur.left;
                }else{
                    pre.right = null;
                    stack.add(new TreeNode(cur.val));
                    cur = cur.right;
                }
            }
        }
    }

    public int next() {
        return stack.pop().val;
    }

    public boolean hasNext() {
        return !stack.isEmpty();
    }
}

方式二(中序遍历)：

class BSTIterator {
    private List<Integer> inOrder;
    private int index;

    public BSTIterator(TreeNode root) {
        inOrder = new ArrayList<>();
        index  = -1;
        inOrderTravel(root);
    }

    private void inOrderTravel(TreeNode root){
        if(root == null)
            return;
        inOrderTravel(root.left);
        this.inOrder.add(root.val);
        inOrderTravel(root.right);
    }
    
    public int next() {
        return inOrder.get(++index);
    }
    
    public boolean hasNext() {
        return inOrder.size() - index - 1 > 0;
    }
}

方式三(只保存左子树节点)：

class BSTIterator {
    private Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack  = new Stack<>();
        this.leftMostInOrder(root);
    }
    private void leftMostInOrder(TreeNode root){
        while(root != null){
            stack.push(root);
            root = root.left;
        }
    }
    
    public int next() {
        TreeNode node = stack.pop();
        if(node.right != null)
            leftMostInOrder(node.right);
        return node.val;
    }
    
    public boolean hasNext() {
        return !stack.isEmpty();
    }
}

参考：

https://leetcode-cn.com/problems/binary-search-tree-iterator/solution/er-cha-sou-suo-shu-die-dai-qi-by-leetcode/