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# 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.

## Thoughts

检验给定BST是否合法。BST进行中序遍历一定是升序的，利用此性质检查当前结点值是否>上个结点。

## Code

```
class Solution {
    TreeNode prev = null;
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        if (!isValidBST(root.left) || prev != null && prev.val >= root.val) {
            return false;
        }
        prev = root;
        return isValidBST(root.right);
    }
}
```

## Analysis

做题耗时 5min

时间复杂度同中序遍历，O(n)

## Ver.2

```
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private void pushLeft(Stack<TreeNode> stack, TreeNode node) {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }

    public boolean isValidBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        pushLeft(stack, root);
        TreeNode pre = null;
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (pre != null && pre.val >= node.val) {
                return false;
            }
            pre = node;
            pushLeft(stack, node.right);
        }

        return true;
    }
}
```