173. Binary Search Tree Iterator

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 inaverage.

Example

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

   10
 /    \
1      11
 \       \
  6       12

分析：

把所有的node存在ArrayList，然后用一个指针指向第一个node.

public class BSTIterator {
    Stack<TreeNode> stack;
 
    public BSTIterator(TreeNode root) {
        stack = new Stack<>();
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }

    public boolean hasNext() {
        return !stack.isEmpty();
    }
    
    public int next() {
        if (!hasNext()) return -1;
        
        TreeNode temp = stack.pop();
        TreeNode node = temp.right;
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
        return temp.val;
    }
}