Question

Formatted question description: https://leetcode.ca/all/173.html

 173	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.

 Credits:
 Special thanks to @ts for adding this problem and creating all test cases.

 @tag-tree

Algorithm

The non-recursive form of the in-order traversal of the binary tree requires an additional definition of a stack to assist. The building rule of the binary search tree is left<root<right, and the in-order traversal can extract all nodes from small to large.

Code

Java

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Stack;

public class Binary_Search_Tree_Iterator {

    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class BSTIterator {

        Stack<TreeNode> sk;

        // @note:@memorize: important feature, min is always going left branch to leaf
        public BSTIterator(TreeNode root) {
            sk = new Stack<>();

            // all the way to leftmost leaf
            while (root != null) {
                sk.push(root);
                root = root.left;
            }

        }

        /** @return whether we have a next smallest number */
        public boolean hasNext() {
            return !sk.isEmpty();
        }

        /** @return the next smallest number */
        public int next() {
            TreeNode minNode = sk.pop();

            TreeNode current = minNode;
            // update stack, possible next time min is frmo its right-then-left branch
            if (current.right != null) {
                current = current.right;

                // same logic as in constructor
                while (current != null) {
                    sk.push(current.left);
                    current = current.left;
                }
            }

            return minNode.val;
        }
    }

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */
}

// OJ: https://leetcode.com/problems/binary-search-tree-iterator/
// Time: O(1) amortized
// Space: O(H)
class BSTIterator {
private:
    stack<TreeNode*> s;
    void pushNodes(TreeNode *node) {
        while (node) {
            s.push(node);
            node = node->left;
        }
    }
public:
    BSTIterator(TreeNode* root) {
        pushNodes(root);
    }
    int next() {
        auto node = s.top();
        s.pop();
        pushNodes(node->right);
        return node->val;
    }
    bool hasNext() {
        return s.size();
    }
};

# Definition for a  binary tree node
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class BSTIterator(object):
  def __init__(self, root):
    """
    :type root: TreeNode
    """
    self.p = None
    self.stack = []
    if root:
      self.stack.append((1, root))

  def hasNext(self):
    """
    :rtype: bool
    """
    return len(self.stack) > 0

  def next(self):
    """
    :rtype: int
    """
    stack = self.stack
    while stack:
      p = stack.pop()
      if not p[1]:
        continue
      if p[0] == 0:
        return p[1].val
      else:
        l = []
        if p[1].right:
          l.append((1, p[1].right))
        l.append((0, p[1]))
        if p[1].left:
          l.append((1, p[1].left))
        stack.extend(l)

# Your BSTIterator will be called like this:
# i, v = BSTIterator(root), []
# while i.hasNext(): v.append(i.next())

173 - Binary Search Tree Iterator

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