173. Binary Search Tree Iterator

Problem

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.

Solution (Java)

/**
 * 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;
 *     }
 * }
 */
class BSTIterator {

    private TreeNode node;

    public BSTIterator(TreeNode root) {
        node = root;
    }

    public int next() {
        int res = -1;
        while (node != null) {
            if (node.left != null) {
                TreeNode rightMost = node.left;
                while (rightMost.right != null) {
                    rightMost = rightMost.right;
                }
                rightMost.right = node;
                TreeNode temp = node.left;
                node.left = null;
                node = temp;
            } else {
                res = node.val;
                node = node.right;
                return res;
            }
        }
        return res;
    }

    public boolean hasNext() {
        return node != null;
    }
}

/**
 * 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();
 */

Solution (Javascript)

/**
 * Definition for binary tree
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */

/**
 * @constructor
 * @param {TreeNode} root - root of the binary search tree
 */
var BSTIterator = function(root) {
  this.stack = [];
  this.pushAll(root);
};


/**
 * @this BSTIterator
 * @returns {boolean} - whether we have a next smallest number
 */
BSTIterator.prototype.hasNext = function() {
  return this.stack.length !== 0;
};

/**
 * @this BSTIterator
 * @returns {number} - the next smallest number
 */
BSTIterator.prototype.next = function() {
  var node = this.stack.pop();
  this.pushAll(node.right);
  return node.val;
};

/**
 * Your BSTIterator will be called like this:
 * var i = new BSTIterator(root), a = [];
 * while (i.hasNext()) a.push(i.next());
*/

BSTIterator.prototype.pushAll = function (node) {
  while (node) {
    this.stack.push(node);
    node = node.left;
  }
};

Explain:

nope.

Complexity:

• Time complexity :
• Space complexity :