# 98 Validate Binary Search Tree

## 98. Validate Binary Search Tree

## 1. Question

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:**

```
    2
   / \
  1   3
```

Binary tree`[2,1,3]`, return true.

**Example 2:**

```
    1
   / \
  2   3
```

Binary tree`[1,2,3]`, return false.

## 2. Implementation

(1) Inorder Traversal

```java
class Solution {
    public boolean isValidBST(TreeNode root) {
        TreeNode curNode = root, preNode = null;
        Stack<TreeNode> stack = new Stack<>();

        while (curNode != null || !stack.isEmpty()) {
            if (curNode != null) {
                stack.push(curNode);
                curNode = curNode.left;
            }
            else {
                curNode = stack.pop();
                if (preNode != null && preNode.val >= curNode.val) {
                    return false;
                }
                preNode = curNode;
                curNode = curNode.right;
            }
        }
        return true;
    }
}
```

(2) Binary Search Tree Property

```java
class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    public boolean isValidBST(TreeNode node, long min, long max) {
        if (node == null) {
            return true;
        }

        if (node.val <= min || node.val >= max) {
            return false;
        }
        return isValidBST(node.left, min, node.val) && isValidBST(node.right, node.val, max);
    }
}
```

## 3. Time & Space Complexity

Inorder Traversal: 时间复杂度:O(n), 空间复杂度: O(h)

Binary Search Tree Property: 时间复杂度: O(n), 空间复杂度: O(h)