I have this algorithm to sum binary tree branches from leftmost branch to rightmost one, so the solution is an array of sums:

class BinaryTree {
  constructor(value) {
    this.value = value;
    this.left = null;
    this.right = null;

function branchSums(root) {
  let sums = [];
    _doSum(root, 0, sums);
    return sums;

function _doSum(root, currentSum, sums) {
    if (!root) {
    if (!root.left && !root.right) {
        sums.push(currentSum + root.value);
    } else {
        _doSum(root.left, currentSum + root.value, sums);
        _doSum(root.right, currentSum + root.value, sums);

It's working fine, but now I'm trying to solve the same problem in an iterative fashion.

I have tried this:

function branchSums(root) {
  let stack = [];
    let sums = [];
    let sum = 0;
    while (stack.length > 0) {
        const node = stack.pop();
        sum += node.value;
        if (!node.right && !node.left) {
            sum -= node.value;
        if (node.right) {
        if (node.left) {
    return sums;

But it doesn't work. After debugging, I have noticed that the error is because I'm only decreasing sum when I reach a leaf node, after pushing to sums. But I need to do it also when depth changes. I don't know how to explain it.

Is it possible to wirte an iterative solution to this problem?

  • 1
    $\begingroup$ It certainly is possible, as every recursive algorithm can be written as an iterative one, using a stack as you do. It gets more complicated when you have a >1 branching factor in the recursive version. This might be helpful: stackoverflow.com/questions/159590/… (take a look at the 2nd and 3rd answers in particular) $\endgroup$ – CSSTUDENT Nov 25 '20 at 18:39

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