# Construct binary tree from inorder and preorder traversal

console.clear();

function cons(inorder, preorder) {
if (isEmpty(inorder) || isEmpty(preorder)) return null

if (size(inorder) === 1 && size(preorder) === 1)
return {val: inorder, l: null, r: null};

return {
val: preorder,
l: cons(take(inorder, leftSize), take(tail(preorder), leftSize)),
r: cons(drop(inorder, leftSize + 1), drop(tail(preorder), leftSize))
};
}

return xs;
}

function size(xs) {
return xs.length;
}

function isEmpty(xs) {
return xs.length === 0;
}

function take(xs, i) {
return xs.slice(0, i);
}

function drop(xs, i) {
return xs.slice(i);
}

function tail(xs) {
return xs.slice(1);
}

var inorder  = [1, 2, 3, 4, 5, 6];
var preorder = [4, 2, 1, 3, 5, 6];

var tree = cons(inorder, preorder);
console.log(tree.r.val === 5);
console.log(tree.l.val === 2);
console.log(tree.r.r.val === 6);
console.log(tree.l.l.val === 1);
console.log(tree.l.r.val === 3);

• Could you please include some text with what this code is supposed to do? – Sumurai8 Aug 22 '16 at 15:59
• @Sumurai8 the question is very self explanatory. – CodeYogi Aug 22 '16 at 16:14
• The title of a question is never the first sentence of the question. The question body only contains code. I find that it does not make for a good question if there is no text at all in the question. I am not sure what you are trying to accomplish with having both an inorder and preorder traversal. – Sumurai8 Aug 22 '16 at 17:20

When inorder and preorder both have a single element, it must be the same value, and the left and right parts in inorder are empty. Therefore the special treatment for size(inorder) === 1 && size(preorder) === 1 is unnecessary, you can simply drop that condition and the algorithm remains correct.
l is the worst variable name ever. Depending on your font, it may be difficult to discern from 1 or |. In this example, I would spell out left and right for the tree branches, naturally.