Merge BSTs to Create Single BST LeetCode solution in JavaScript

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var canMerge = function (trees) {
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  let Node = {},
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    indeg = {};
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  // traverse the mini trees and put back pointers to their parents, also figure out the indegree of each node
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  let dfs = (node, leftparent = null, rightparent = null) => {
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    if (!node) return;
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    indeg[node.val] =
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      indeg[node.val] || Number(leftparent !== null || rightparent !== null);
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    (node.lp = leftparent), (node.rp = rightparent);
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    dfs(node.left, node, null), dfs(node.right, null, node);
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  };
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  for (let root of trees) (Node[root.val] = root), dfs(root);
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  //there are a lot of potential roots=> no bueno
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  if (
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    Object.values(indeg).reduce((a, b) => a + b) !=
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    Object.keys(indeg).length - 1
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  )
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    return null;
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  //find THE root
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  let bigRoot,
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    timesMerged = 0;
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  for (let root of trees) if (indeg[root.val] === 0) bigRoot = root;
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  // traverse the tree while replacing each leaf that can be replaced
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  let rec = (node = bigRoot) => {
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    if (!node) return;
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    if (!node.left && !node.right) {
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      let toadd = Node[node.val];
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      Node[node.val] = undefined; //invalidating the trees you already used
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      if (toadd === undefined) return;
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      //make the change
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      if (node.lp === null && node.rp === null) return;
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      else if (node.lp !== null) node.lp.left = toadd;
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      else node.rp.right = toadd;
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      timesMerged++;
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      rec(toadd);
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    } else rec(node.left), rec(node.right);
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  };
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  rec();
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  var isValidBST = function (node, l = -Infinity, r = Infinity) {
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    //l and r are the limits node.val should be within
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    if (!node) return true;
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    if (node.val < l || node.val > r) return false;
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    return (
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      isValidBST(node.left, l, node.val - 1) &&
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      isValidBST(node.right, node.val + 1, r)
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    );
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  };
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  //check if every item was used and if the result bst is valid
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  return !isValidBST(bigRoot) || timesMerged !== trees.length - 1
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    ? null
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    : bigRoot;
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};