Recover Binary Search Tree

Problem statement

You are given the root of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.

Example 1:

Input: root = [1,3,null,null,2]Output: [3,1,null,null,2]Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.

Example 2:

Input: root = [3,1,4,null,null,2]Output: [2,1,4,null,null,3]Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.

Constraints:

• The number of nodes in the tree is in the range [2, 1000].
• -231 <= Node.val <= 231 - 1

Follow up: A solution using O(n) space is pretty straight-forward. Could you devise a constant O(1) space solution?

My solution

/**
 * 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 {void} Do not return anything, modify root in-place instead.
 */
var recoverTree = function(root) {
    let prev = null;
    let small = null;
    let big = null;
    
    function dfs(node) {
        if (!node) {
            return;
        }
        
        dfs(node.left)
        
        if (prev !== null && prev.val > node.val) {
            small = node;
            if (!big) {
                big = prev;
            } else {
                return
            }
        }
        
        prev = node;
        
        dfs(node.right)
    }
    
    
    dfs(root)
    
    // console.log(small, big)
    const temp = big.val;
    big.val = small.val;
    small.val = temp;
    
    // [big.val, small.val] = [small.val, big.val]
};