173. 二叉搜索树迭代器

题目链接

https://leetcode-cn.com/problems/binary-search-tree-iterator/

 

做法1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
/**
     * 利用二叉搜索树的性质, 每次弹出左子树的左节点，这样子可以保证每次弹出都是最小的数
     */

    class BSTIterator {
        private Stack<TreeNode> stack = new Stack<>();
        
        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 treeNode = stack.pop();
            pushAll(treeNode.right);
            return treeNode.val;
        }

       

        private void pushAll(TreeNode treeNode){
            for (; treeNode != null; stack.push(treeNode), treeNode = treeNode.left);
        }
    }
/**
 * 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();
 */

做法2

/**
     * 做法2 利用中序遍历，从小到大取出所有节点
     */
    class BSTIterator {
     private  Stack<TreeNode> stack = new Stack<>();
    public BSTIterator(TreeNode root) {
        while (root != null){
            stack.push(root);
            if (root.left != null){
                root = root.left;
            }
            else
                break;
        }
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode root =stack.pop();
        TreeNode res = root;
        if (res.right != null){
            res = res.right;
            while (res != null){
                stack.push(res);
                if (res.left != null)
                    res = res.left;
                else
                    break;
            }
        }
        return root.val;
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }
}