All Nodes Distance K in Binary Tree LeetCode solution in JavaScript

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/**
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 * Definition for a binary tree node.
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 * function TreeNode(val) {
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 * this.val = val;
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 * this.left = this.right = null;
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 * }
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 */
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/**
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 * @param {TreeNode} root
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 * @param {TreeNode} target
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 * @param {number} k
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 * @return {number[]}
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 */
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var distanceK = function (root, target, k) {
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  if (k === 0) return [target.val];
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  let res = [];
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  const helper = (node, k, bool) => {
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    if (node === null) return [false, 0];
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    if (k === 0 && node.val !== target.val) {
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      res.push(node.val);
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      return null;
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    }
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    if (bool) {
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      helper(node.left, k - 1, bool);
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      helper(node.right, k - 1, bool);
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      return null;
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    }
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    if (node.val === target.val) {
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      //Here nodes of k distance are at bottom
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      helper(node.left, k - 1, true);
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      helper(node.right, k - 1, true);
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      return [true, 1];
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    } else {
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      let [l, d] = helper(node.left, k, false);
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      //Found target node on left side, now get nodes of k distance from right side as well
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      if (l === true) {
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        if (k - d === 0)
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          //for current node
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          res.push(node.val);
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        helper(node.right, k - d - 1, true);
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        return [true, d + 1];
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      }
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      let [r, d1] = helper(node.right, k, false);
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      if (r === true) {
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        if (k - d1 === 0) res.push(node.val);
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        helper(node.left, k - d1 - 1, true);
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        return [true, d1 + 1];
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      }
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      return [false, 0];
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    }
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  };
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  helper(root, k, false);
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  return res;
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};