700. Search in a Binary Search Tree

You are given the root of a binary search tree (BST) and an integer val.
Find the node in the BST that the node’s value equals val and return the subtree rooted with that node. If such a node does not exist, return null.

Example 1:

Input: root = [4,2,7,1,3], val = 2 Output: [2,1,3]
Example 2:

Input: root = [4,2,7,1,3], val = 5 Output: []

Constraints:

• The number of nodes in the tree is in the range [1, 5000].
• 1 <= Node.val <= 107
• root is a binary search tree.
• 1 <= val <= 107

Runtime: 92 ms, faster than 84.88% of JavaScript online submissions for Search in a Binary Search Tree.
Memory Usage: 45.2 MB, less than 33.90% of JavaScript online submissions for Search in a Binary Search Tree.

/**
 * 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
 * @param {number} val
 * @return {TreeNode}
 */
var searchBST = function(root, val) {
    if (!root) {
        return null;
    }
    if (root.val === val) {
        return root;
    }
    if (root.val > val) {
        return searchBST(root.left, val);
    }
    return searchBST(root.right, val);
};

Runtime: 88 ms, faster than 92.84% of JavaScript online submissions for Search in a Binary Search Tree.
Memory Usage: 45.6 MB, less than 8.59% of JavaScript online submissions for Search in a Binary Search Tree.

/**
 * 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
 * @param {number} val
 * @return {TreeNode}
 */
var searchBST = function(root, val) {
    if (!root) {
        return null;
    }
    let node = root;
    while(node) {
        if (node.val === val) {
            return node;
        }
        if (node.val > val) {
            node = node.left;
        } else {
            node = node.right;
        }
    }
    return null;
};