构造哈夫曼树

哈弗曼树中没有度为一的节点，是标准的二叉树，所以有n个叶子结点时，需要一个长度为2n-1的一维数组存储哈弗曼树的结点。

(1)、n个叶子节点只有weight权值，处理非叶子节点，从ht[i]（ht[1]~ht[n-1]）中找到ht[i].weight最小的两个节点ht[s1]和ht[s2]，这就是Select(int n,int &s1,int & s2,HTNode *ht)函数完成的功能。

(2)、调用select函数，并将ht[s1]和ht[s2]作为ht[l]的左右子树，即ht[s1]和ht[s2]双亲节点为ht[l]，新的根节点的权值为其左右子树权值之和， ht[l].weight=ht[s1].weight+ht[s2].weight

-Huffman tree structure tree哈弗曼no degree 1 nodes, is a standard binary tree, there is a leaf node n, the need for a length of 2n-1 one-dimensional array of storage哈弗曼tree nodes. (1), n a leaf node, only the right to weight the value of treatment of non-leaf node, from ht [i] (ht [1] ~ ht [n-1]) found in ht [i]. Weight minimum of two nodes ht [s1] and ht [s2], This is the Select (int n, int