# [LeetCode] BST Series #98 #99 #109 #173 #270

## 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:

``````Input:
2
/ \
1   3
Output: true``````

Example 2:

``````    5
/ \
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) {
stack.push(root);
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) {
stack.push(cur);
cur = cur.left;
}
}
return node.val;
}
}``````

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