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())
    
    

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