理想情况下，您应该自己按顺序遍历节点。参见：http ://en.wikipedia.org/wiki/In-order_traversal#Inorder_Traversal
但是，此功能应该为您提供您正在寻找的确切功能：
function convertPreOrderArrayToInOrderString(elements)
{
var cursor = 0;
// While all the elements aren't merged together
while (elements.length > 1)
{
// Skip to the nextleaf
while (elements[cursor] === "AND" || elements[cursor] === "OR") cursor++;
// If we made all the matches we can at this depth, go to the next one
if (typeof elements[cursor+1] === "undefined")
{
cursor = 0;
continue;
}
// The relevant pieces to combining
var previous = elements[cursor-1];
var current = elements[cursor];
var next = elements[cursor+1];
// Create a string from the pieces
var tmpStr = current + " " + previous;
// How many we've combined
var combined = 2;
// If the third piece is complete, we can join it too
if (next !== "AND" && next != "OR")
{
// Tack the third piece on
tmpStr += " " + next;
combined++;
}
// Remove the individual elements and replace with the combined ones
var newElements = elements.slice(0, cursor-1);
newElements.push(tmpStr);
newElements = newElements.concat(elements.slice(cursor - 1 + combined));
// From now on, opreate on the new array with combined elements
elements = newElements;
}
// There should be one element left
return elements[0];
}
var expression = ["AND","OR","AND","a","f","OR","e","b","OR","c","d"];
console.log(convertPreOrderArrayToInOrderString(expression));
// Gives: "a AND f OR e OR b AND c OR d"
对应的小提琴：http: //jsfiddle.net/DvZj4/4/
不过，在更一般的情况下，您可能更喜欢这样做：
function convertPreOrderToInOrder(elements)
{
// Very simple sanity checks
if (elements.length === 1 && (elements[0] === "AND" || elements[0] === "OR"))
{
console.error("Cannot construct compound expression with boolean connectors without values");
return;
}
else if (elements.length === 1)
{
// Only 1 items, just return it
return elements[0];
}
else if (elements.length > 1 && elements[0] !== "AND" && elements[0] !== "OR")
{
console.error("Cannot construct compound expression without boolean connectors");
return;
}
// Convert back to a tree from the preorder, then flatten again in order
return flattenTreeInOrder(getTreeFromPreOrderArray(elements));
}
// Reconstructs a tree from a pre-ordered array of elements
// Assumes your array of elements can construct a properly
// balanced tree. If it can't, this will get screwy
function getTreeFromPreOrderArray(elements)
{
var i, el, root = {}, current = root;
// Iterate over every element
for (i in elements)
{
el = elements[i];
// Pretty much just a root check
if ( (el === "AND" || el === "OR") && !current.connector)
{
// Make this a root
current.connector = el;
}
// Not a root, can we go left?
else if (!current.left)
{
// Branch left
if (el === "AND" || el === "OR")
{
current.left = {connector: el};
current.left.parent = current;
current = current.left;
}
// Left left
else
{
current.left = el;
current.left.parent = current;
}
}
// Not a root, can't go left. Try to the right.
else
{
// Branch right
if (el === "AND" || el === "OR")
{
current.right = {connector: el};
current.right.parent = current;
current = current.right;
}
// Leaf Right
else
{
current.right = el;
current.right.parent = current;
// Tricky bit: If we put a leaf on the right, we need
// to go back up to the nearest parent who doesn't have
// right child;
while (current.parent && current.right)
{
current = current.parent;
}
}
}
}
return root;
}
// Flatten a tree into an array using in-order traversal
// No recursion, uses stack-based traversal
// Returns an array of strings
function flattenTreeInOrder(root)
{
var stack = [];
var flat = [];
var current = root;
// We need to keep going as long as there's either something in the
// stack, or we're looking at a node.
while (stack.length > 0 || current)
{
// If the looked at node is defined
// This will help us tell if we've gone past a leaf
if (current)
{
// Don't care during in order traversal. Just push this on
// and move to the left
stack.push(current);
current = current.left;
}
else
{
// Starting to go back up, but the current is undefined
// So, pop first, then start capturing in order elements
current = stack.pop();
flat.push(current.connector || current);
current = current.right;
}
}
return flat;
}
var expression = ["AND","OR","AND","a","f","OR","e","b","OR","c","d"];
console.log(convertPreOrderToInOrder(expression).join(" "));
// Gives: "a AND f OR e OR b AND c OR d"
对应小提琴：http: //jsfiddle.net/keR5j/
不能复制 htmlid元素
此外，您不能像在结构中那样使用 id。有多个相同的实例id，即每个规范的两个节点id="OR"是无效的 HTML：http: //www.w3.org/TR/html401/struct/global.html#h-7.5.2
id = 名称 [CS]
此属性为元素分配名称。此名称在文档中必须是唯一的。