f＃访问树的根元素(f# access root element of a tree)

 我在F＃中有一个规范树，即通过声明 
 
type binaryTree = 
    | Leaf
    | Node of binaryTree * float * binaryTree
 
 然后使用递归函数来创建树 
 
let rec makeTree tree element = 
    match element, tree with
        | x, Leaf -> Node(Leaf,x,Leaf)
        | x, Node(l,y,r) -> Node(l,y, (makeTree r x))
 
 这一切都很好。 现在我想对树进行排序，以便在每个节点上，节点的值小于其所有子节点的值。 我可以想象这样做。 但是，我想采用树的第一个元素。 也就是说，我想将树视为队列。 我在树上看到的唯一例子是使用高阶函数来对树做一些事情，但是当我已经对它进行排序时，这似乎是一种浪费。 
 
 如何访问此树的根节点？ 

I have a canonical tree in F#, i.e by declaring
 
type binaryTree = 
    | Leaf
    | Node of binaryTree * float * binaryTree
 
and then using a recursive function to make the tree
 
let rec makeTree tree element = 
    match element, tree with
        | x, Leaf -> Node(Leaf,x,Leaf)
        | x, Node(l,y,r) -> Node(l,y, (makeTree r x))
 
This is all fine. Now I want to sort the tree so that at each node, the value of the node is smaller than the value of all its children. I can imagine doing this. However, I want to then take the first element of the tree. That is, I want to treat the tree like a queue. The only examples I have seen with trees use higher-order functions to do something with the tree, but this seems like a waste when I have already sorted it.
 
How can I access the root node of this tree?

原文：https://stackoverflow.com/questions/5817777