取两个HTML节点最近的公共父节点

gistfile1.js

function getCommonParent(el1,el2){

    var parents1 = [];
    var el = el1;
    while(el) {
        parents1.unshift(el);
        el = el.parentNode;
    }

    var parents2 = [];
    var el = el2;
    while(el) {
        parents2.unshift(el);
        el = el.parentNode;
    }

    var i = 0;

    while(i<parents1.length && i<parents2.length && parents1[i+1] == parents2[i+1])
        i++;

    return parents1[i];
}

function getCommonParent(el1,el2) {
    if (!el1 || !el2)
        return null;

    var commonParent = null;
    var nodes = [el1, el2];
    for (var i=0; i<nodes.length; ++i) {
        var node = nodes[i];
        if (node.gcpVisited) {
            commonParent = node;
            break;
        }
        node.gcpVisited = true;
        if (node = node.parentNode) {
            nodes.push(node);
        }
    }

    while (node = nodes.pop()) {
        delete node.gcpVisited;
    }
    return commonParent;
}

两边一起来，如果离共同父类距离相近时则查询次数较少，但有设flag和清flag消耗，综合实力就不清楚了。

我也来一段吧

function getCommonParent(a,b){
    if(a.sourceIndex){ //for IE
        var sib = b.sourceIndex, sia = a.sourceIndex;
        if(sib < sia){
            return getCommonParent(b,a);
        }
        while((sib > sia) && (b = b.parentNode)){
            sib = b.sourceIndex;
        }
        return b && b.contains(a) ? b : null;
    }else{ //Not IE
        //...参考chaoren1641
    }
}

@cuixiping 真是各种高人啊...... 长见识了

再来一段使用的标准compareDocumentPosition的，把代码修改得简洁了一点。

function getCommonParent(a,b){
    var c = a.compareDocumentPosition(b);
    if(c & 8){
        return b.parentNode;
    }else if(c & 16){
        return a.parentNode;
    }else if(c & 6){
        while(!(8 & a.compareDocumentPosition(b=b.parentNode)));
        return b;
    }
    return null;
}