Binary Search Tree Iterator

lc 173

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.

这个题关键有个cur 的全局变量node，时刻知道现在哪里了。然后有个stack全局变量，inorder 走就行了。

public class BSTIterator {
    private Stack<TreeNode> stack = new Stack<>();
    private TreeNode cur;

    public BSTIterator(TreeNode root) {
        cur = root;


    }

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

    /** @return the next smallest number */
    public int next() {
        while(cur != null){
            stack.push(cur);
            cur = cur.left;
        }
        TreeNode temp = stack.pop();
        cur = temp.right;
        return temp.val;
    }
}