Binary Search Tree Iterator

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.

思路一：

使用inorder traversal把tree转化为arraylist。

public class BSTIterator {
        
        ArrayList<TreeNode> list;
        int index;

        public BSTIterator(TreeNode root) {
            list  = new ArrayList<TreeNode>();
            iterator(root, list);
            index = 0;
        }
        
        public void iterator(TreeNode root, ArrayList<TreeNode> list) {
            if(root != null) {
                iterator(root.left, list);
                list.add(root);
                iterator(root.right, list);
            }
        }

        /** @return whether we have a next smallest number */
        public boolean hasNext() {
            if(index < list.size()) {
                return true;
            }
            return false;
        }

        /** @return the next smallest number */
        public int next() {
            return list.get(index++).val;
        }
    }

 思路二：

将iterator的实现改为迭代。

public void iterator(TreeNode root, ArrayList<TreeNode> list) {
            Stack<TreeNode> stack = new Stack<TreeNode>();
            TreeNode node = root;
            while(!stack.isEmpty() || node != null) {
                if(node != null) {
                    stack.push(node);
                    node = node.left;
                } else {
                    node  = stack.pop();
                    list.add(node);
                    node = node.right;
                }
            }
        }