【数据结构】求节点的哈夫曼的带权路径长度

题目来源：北航14级6系数据结构作业

【问题描述】
已知输入一串正整数，正整数之间用空格键分开，请建立一个哈夫曼树，以输入的数字为叶节点，求这棵哈夫曼树的带权路径长度。

【输入形式】
首先输入正整数的个数，然后接下来为接下来的正整数，正整数个数不超过10个

【输出形式】
输出相应的权值

【样例输入】
5 4 5 6 7 8

【样例输出】
69

求哈夫曼树的算法用的是递归，然而并不会写非递归的，还要花时间研究一下，这次没有看清楚输入的要求，走了，不少的弯路，下次看题目一定要认真啊

#include <stdio.h>
#include <stdlib.h>

#define MAxnuM 20

typedef struct BTNode {
    int data;
    struct BTNode *lchild,*rchild;
}BTNode,*BTree;

typedef struct Node {
    BTree bt;
    struct Node *link;
} Node,*Forest;


void insertForestNode ( BTree t,Forest fr,Forest head);
void destoryBTree ( BTree t );
void getWPL( BTree t,int depth,int *WPL);


int main() {

    int num;
    int n;
    int i;
    Forest head,top,fr,p,q;
    BTree t,root,t1,t2;
    int WPL = 0;

    head = (Forest)malloc(sizeof(Node));
    head->link = NULL;

    scanf("%d",&n);
    //用链式队列存森林
    i = 1;
    while( i <= n ) {

        scanf("%d",&num);

        t = (BTree)malloc(sizeof(BTNode));
        t->data = num;
        t->lchild = t->rchild = NULL;

        fr = (Forest)malloc(sizeof(Node));
        fr->bt = t;
        fr->link = NULL;

        //将元素从大到小排列
        if( head->link == NULL) {
            head->link= fr;
            //top = fr;
        }

        else
            insertForestNode(t,head);

        i++;
    }

    //构造哈夫曼树
    while( head->link != NULL ) {

        t = (BTree)malloc(sizeof(BTNode));
        q = (Forest)malloc(sizeof(Node));
        q->bt = t;
        q->link = NULL;

        t1 = head->link->bt;
        p = head->link;
        head->link = p->link;
        t->lchild = t1;
        free(p);

        t2 = head->link->bt;
        p = head->link;
        head->link = p->link;
        t->rchild = t2;
        free(p);

        t->data = t1->data + t2->data;
        if(head->link != NULL)
            insertForestNode(t,q,head);
    }
    free(head);

    getWPL(t,&WPL);
    printf("%d",WPL);

    destoryBTree(t);
    t = NULL;

    return 0;
}

void insertForestNode ( BTree t,Forest head ) {

    Forest p = head->link,q;

    while ( p != NULL ) {

        if( fr->bt->data > p->bt->data) {
            q = p;
            p = p->link;
        }

        else {
            fr->link = p;

            if( p == head->link)
                head->link = fr;
            else
                q->link = fr;

            break;
        }

    }

    if( p == NULL)
      q->link = fr;

}

//int getWPL ( BTree t) {
//
//    int depth = 0;
//    int top = -1;
//    int WPL = 0;
//    BTree Stack[MAxnuM];
//    BTree p;
//
//    p = t;
////    do{
////
////        while( p != NULL ) {
////            Stack[++top] = p;
////            p = p->lchild;
////            depth++;
////        }
////
////        p = Stack[top--];
////        WPL = WPL + p->data*(--depth);
////        p =
////
////    } while ( p != NULL || top != -1 );
//    do{
//
//        if( p->lchild != NULL && p->rchild != NULL ) {
//            Stack[++top] = p;
//            depth++;
//            p = p->lchild;
//
//        }
//
//        else {
//            WPL = WPL + p->data*(depth);
//            p = Stack[top]->rchild;
//            top--;
//
//        }
//    } while( top != -1);
//
//    return WPL;
//}


void getWPL( BTree t,int *WPL) {
    if( t->lchild == NULL && t->rchild == NULL) {
        (*WPL) += depth*t->data;
    }

    else{
        getWPL(t->lchild,depth+1,WPL);
        getWPL(t->rchild,WPL);
    }

}

//二叉树的销毁
void destoryBTree ( BTree t ) {

    if( t != NULL ) {
        destoryBTree(t->lchild);
        destoryBTree(t->rchild);
        free(t);
    }

}

经过某同学指点，利用了队列和按层次遍历，这下就可以有非递归算法了

先定义一个结构存有节点地址和其高度

struct depthBuffmanTree {
    BTree bt;
    int depth;
};

接着是

//非递归算法
int getWPL2(BTree t) {
    struct depthBuffmanTree queue[MAxnuM];
    struct depthBuffmanTree p;
    int front = -1;
    int rear = 0;
    int WPL = 0;

    queue[0].bt = t;
    queue[0].depth = 0;

    while( front < rear ) {
        p = queue[++front];

        if(p.bt->lchild == NULL && p.bt->rchild == NULL) {
            WPL += p.bt->data*p.depth;
        }

        if( p.bt->lchild != NULL ) {
            queue[++rear].bt = p.bt->lchild;
            queue[rear].depth = p.depth + 1;
        }

        if( p.bt->rchild != NULL ) {
            queue[++rear].bt = p.bt->rchild;
            queue[rear].depth = p.depth + 1;
        }

    }

    return WPL;

}

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