173. Binary Search Tree Iterator (stack)

time: O(n), n is node nums, to constuct iterator

space: O(n), use stack, there are n nodes number add to stack

next(): O(1), average, somtimes need add left nodes

hasNext(): O(1),

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 
 因為 inorder 都是有序的, 所以從最小開始（ leftmost
 一路 push 到 stack, 這樣取出時, 會從最小開始取出
 
 
   8  -> root
  /
 2  
/ \    
1  4   => [1, 2, 8]

from root, do leftmost to push into stack
stack
1  => pop1
2
8

2 => pop 2 => 4
8


4
8
 
 */
class BSTIterator {
    private Deque<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        stack = new ArrayDeque<>();
        findMostLeft(root); // add left all first
    }
    private void findMostLeft(TreeNode node) { // add left all first
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }
    
    public int next() {
        TreeNode node = stack.pop();
        if (node.right != null) { // if have right subtree
            findMostLeft(node.right); // add right node's left all first
        }
        return node.val;
    }
    
    public boolean hasNext() {
        return !stack.isEmpty();
    }
}

/**
 * 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();
 */