98. Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
Example 1:

   / \
  1   3
Output: true

Example 2:

   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

Solution #98

class Solution {
    public boolean isValidBST(TreeNode root) {
        return helper(root, null, null);
    private boolean helper(TreeNode root, Integer min, Integer max) {
        if (root == null) return true;
        if (min != null && root.val <= min) return false;
        if (max != null && root.val >= max) return false;
        return helper(root.left, min, root.val) && helper(root.right, root.val, max);

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.

Solution #173

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

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

    /** @return the next smallest number */
    public int next() {
        TreeNode node = stack.pop();
        if (node.right != null) {
            TreeNode cur = node.right;
            while (cur != null) {
                cur = cur.left;
        return node.val;

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