Given the `root` of a binary tree, determine if it is a valid binary search tree (BST).

valid 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: root = [2,1,3]
Output: true
```

Example 2:

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

Constraints:

• The number of nodes in the tree is in the range `[1, 104]`.
• `-231 <= Node.val <= 231 - 1`

Idea:

Use Binary Tree Inorder Traveral Template

Solution 1:

``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isValidBST = function(root) {
// use inorder traversal template (2 whiles, 里左外右)
let preVal = -Infinity;
let stack = [];

while (root !== null || stack.length !== 0) {
while (root !== null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if (root.val <= preVal) return false;
preVal = root.val;
root = root.right;
}
return true;
};``````

Solution 2: (Recursion)

``````var isValidBST = function(root) {
return helper(root, null, null);
}

function helper(node, low, high) {
if (node === null) return true;
const val = node.val;
if ((low !== null && val <= low) || (high !== null && val >= high))
return false;
return helper(node.right, val, high) && helper(node.left, low, val);
}``````