[JS] Ancestors do-while

/*
Suppose we have some input data describing a graph of relationships between
parents and children over multiple generations. The data is formatted as a
list of (parent, child) pairs, where each individual is assigned a unique
integer identifier.

For example, in this diagram, 6 and 8 have common ancestors of 4 and 14.
         14  13
         |   |
1   2    4   12
 \ /   / | \ /
  3   5  8  9
   \ / \     \
    6   7     11

Write a function that takes the graph, as well as two of the individuals in our
dataset, as its inputs and returns true if and only if they share at least one
ancestor.
*/

parentChildPairs1 = [
  [1, 3], // x -> parent, y -> child
  [2, 3],
  [3, 6],
  [5, 6],
  [5, 7],
  [4, 5],
  [4, 8],
  [4, 9],
  [9, 11],
  [14, 4],
  [13, 12],
  [12, 9],
];

function hasCommonElement(arr1, arr2) {
  return arr1.some((element) => arr2.includes(element));
}

function hasCommonAncestor(inputData, a, b) {
  const getNextLevelAncestors = (invidual) => {
    let parentAncestors = [];

    inputData.forEach((parentChildPair) => {
      const parent = parentChildPair[0];
      const child = parentChildPair[1];
      if (invidual === child) {
        parentAncestors.push(parent);
      }
    });

    return parentAncestors;
  };

  const getAllAncestors = (invidual) => {
    let ancestors = getNextLevelAncestors(invidual);
    let nextAncetors = [];
    let finalAncestors = [];

    do {
      nextAncetors = ancestors.reduce((acc, curr) => {
        const currAncestors = getNextLevelAncestors(curr);
        return [...acc, ...currAncestors];
      }, []);

      ancestors = nextAncetors;
      finalAncestors = [...finalAncestors, ...ancestors];
    } while (nextAncetors.length !== 0);

    return finalAncestors;
  };

  return hasCommonElement(getAllAncestors(a), getAllAncestors(b));
}

console.log(hasCommonAncestor(parentChildPairs1, 5, 6));

// hasCommonAncestor(parentChildPairs1, 3, 8); // => false
// hasCommonAncestor(parentChildPairs1, 5, 8); // => true
// hasCommonAncestor(parentChildPairs1, 6, 8); // => true
// hasCommonAncestor(parentChildPairs1, 6, 9); // => true
// hasCommonAncestor(parentChildPairs1, 1, 3); // => false
// hasCommonAncestor(parentChildPairs1, 3, 1); // => false
// hasCommonAncestor(parentChildPairs1, 7, 11); // => true
// hasCommonAncestor(parentChildPairs1, 6, 5); // => true
// hasCommonAncestor(parentChildPairs1, 5, 6); // => true