Problem

Design an iterator over a binary search tree with the following rules:

Elements are visited in ascending order (i.e. an in-order traversal)

next() and hasNext() queries run in O(1) time in average.

Example
For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]

   10
 /    \
1      11
 \       \
  6       12

思路

• 因为是中序遍历, 所以每次加一个节点, 需要把所有的左子节点push到stack里面
• hasNext()
  • 检查stack是否为空
• next()
  • 取出stack顶元素, 并将该元素的右子树及右子树的所有左子节点加入到stack中
  • 返回顶元素
• addToStack(curt)
  • 将curt以及它的所有左子孙节点都加到stack中

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Example of iterate a tree:
 * BSTIterator iterator = new BSTIterator(root);
 * while (iterator.hasNext()) {
 *    TreeNode node = iterator.next();
 *    do something for node
 * } 
 */

public class BSTIterator {
    //@param root: The root of binary tree.

    Stack<TreeNode> stack;

    private void addToStack(TreeNode root) {
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }

    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        addToStack(root);
    }

    //@return: True if there has next node, or false
    public boolean hasNext() {
        if (!stack.isEmpty()) {
            return true;
        }
        return false;
    }

    //@return: return next node
    public TreeNode next() {
        if (!hasNext()) {
            return null;
        }
        TreeNode cur = stack.pop();
        addToStack(cur.right);
        return cur;
    }
}