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benpickles/common-ancestor.js

Last active August 9, 2020 13:49
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Written as a response to a Stack Overflow question "How to find the nearest common ancestors of two or more nodes?" https://stackoverflow.com/a/5350888/194664
Raw
common-ancestor.js
function parents(node) {
var nodes = [node]
for (; node; node = node.parentNode) {
nodes.unshift(node)
}
return nodes
}
function commonAncestor(node1, node2) {
var parents1 = parents(node1)
var parents2 = parents(node2)
if (parents1[0] != parents2[0]) throw "No common ancestor!"
for (var i = 0; i < parents1.length; i++) {
if (parents1[i] != parents2[i]) return parents1[i - 1]
}
}
Raw
test.html
<!DOCTYPE html>
<html>
<head>
<title>Common Ancestor Tests</title>
<link rel="stylesheet" href="http://code.jquery.com/qunit/qunit-1.10.0.css">
<script src="./common-ancestor.js"></script>
</head>
<body>
<div id="qunit"></div>
<script src="http://code.jquery.com/qunit/qunit-1.10.0.js"></script>
<script src="./test.js"></script>
</body>
</html>
Raw
test.js
function createElement(parent) {
var elem = document.createElement("div")
if (parent) {
parent.appendChild(elem)
}
return elem
}
var root = createElement() // .
var elem1 = createElement(root) // ├── elem1
var elem2 = createElement(elem1) // | ├── elem2
var elem3 = createElement(elem1) // | └── elem3
var elem4 = createElement(elem3) // | └── elem4
var elem5 = createElement(root) // └── elem5
test("same parent", function() {
ok(commonAncestor(elem1, elem5) === root)
})
test("same parent but deeper", function() {
ok(commonAncestor(elem4, elem5) === root)
})
test("direct child of the other", function() {
ok(commonAncestor(elem1, elem2) === elem1)
})
test("grandchild of the other", function() {
ok(commonAncestor(elem1, elem4) === elem1)
})
test("when there's no common ancestor", function() {
var separate = createElement()
var thrown = false
try {
commonAncestor(elem3, separate)
} catch(e) {
thrown = true
}
ok(thrown)
})
@jul-sh
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jul-sh commented Nov 26, 2019 •
edited
Loading

Version that accepts more than two nodes

function getParents (node) {
  // Place the first node into an array
  const nodes = [node]

  // As long as there is a parent node that we have yet to add
  while (nodes[nodes.length - 1].parentNode) {
    // Place that node at the end of the array
    // This unravels the tree into an array with child/parents only.
    nodes.push(node)
  }

  // Reverse the array so it starts from the highest ancestor, all the way down
  // to our node.
  nodes.reverse()

  return nodes
}

function isArrayOfIdenticalItems (array) {
  return array.every(item => item === array[0])
}

function getLastCommonAncestor (nodes) {
  // Get an array whose items are arrays of parents
  const arrayOfParents = nodes.map(node => getParents(node))

  /* Although DOM can go many steps up to
   ** a common parent node, top down will be equal for a common ancestor, as
   ** the difference to the top is only after we break past the common ancestor.
   ** After that, the common node has the same parent as the other node, so it's
   ** the same all the way to the top. (If that doesn't make sense, thing about
   ** it, and maybe draw it out).
   */

  // Ensure we have a common ancestor, it will be the first one if anything
  const topMostAncestors = arrayOfParents.map(parents => parents[0])

  
  if (!isArrayOfIdenticalItems(topMostAncestors)) {
    // We do not have a common ancestor, let's throw
    throw new Error('No common ancestor!')
  }

  // Let's move down the ancestor from the top to the bottom
  for (let index = 1; index < arrayOfParents[0].length; index++) {
    const parentsAtIndex = arrayOfParents.map(parents => parents[index])

    if (isArrayOfIdenticalItems(parentsAtIndex)) {
      // It's a common ancestor, let's continue the loop
    } else {
      // It's not a common ancestor, let's hence the common ancestor must be
      // the one from the previous loop
      return arrayOfParents[0][index - 1]
    }
  }
}

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