Tail Recursive Tree Traversal without Loops

Question

I want to traverse the following tree structure tail recursively without falling back on loops :

const o = {x:0,c:[{x:1,c:[{x:2,c:[{x:3},{x:4,c:[{x:5}]},{x:6}]},{x:7},{x:8}]},{x:9}]};

        0
       / \
      1   9
    / | \ 
   2  7  8
 / | \
3  4  6
   |
   5

The desired result: /0/1/2/3/4/5/6/7/8/9

I guess a closure is required to enable tail recursion. I've tried this so far:

const traverse = o => {
  const nextDepth = (o, index, acc) => {
    const nextBreadth = () => o["c"] && o["c"][index + 1]
     ? nextDepth(o["c"][index + 1], index + 1, acc)
     : acc;

    acc = o["c"]
     ? nextDepth(o["c"][0], index, acc + "/" + o["x"]) // not in tail pos
     : acc + "/" + o["x"];

    return nextBreadth();
  };

  return nextDepth(o, 0, "");
};

traverse(o); // /0/1/2/3/4/5/7/9

The siblings are not traversed properly. How can this be done?

Answer 1

As @Bergi wrote if you manually maintain stack the solution is straightforward.

 const o = {x:0,c:[{x:1,c:[{x:2,c:[{x:3},{x:4,c:[{x:5}]},{x:6}]},{x:7},{x:8}]},{x:9}]} const traverse = g => { const dfs = (stack, head) => (head.c || []).concat(stack) const loop = (acc, stack) => { if (stack.length === 0) { return acc } const [head, ...tail] = stack return loop(`${acc}/${head.x}`, dfs(tail, head)) } return loop('', [g]) } console.log(traverse(o)) console.log(traverse(o) === '/0/1/2/3/4/5/6/7/8/9')