LeetCode 98. Validate Binary Search Tree (javascript)

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

A 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);
}