递归计算二叉树叶子结点数.doc

B12-4题目：编写递归算法，计算二叉树中叶子结点的数目。一、 需求分析二、 概要设计int LeafCount_BiTree(Bitree T)/求二叉树中叶子结点的数目if(!T) return 0; /空树没有叶子else if(!T-lchild&!T-rchild) return 1; /叶子结点else return Leaf_Count(T-lchild)+Leaf_Count(T-rchild);/左子树的叶子数加上右子树的叶子数/LeafCount_BiTree 3 详细设计#include#define MaxSize 100#define MaxWidth 40typedef char elemtype;typedef struct node elemtype data; struct node *left,*right; BTree;void dispstack(BTree*stack,int top);void creatree(BTree * &b,char *str) BTree *stackMaxSize,*p; int top=-1,k,j=0; char ch; b=NULL; ch=strj; while (ch!=0) switch(ch) case (: top+;stacktop=p;k=1;break; case ): top-;break; case ,: k=2;break; default: p=(BTree *)malloc(sizeof(BTree); p-data=ch;p-left=p=-right=NULL; if(b=NULL) b=p; else switch(k) case 1: stacktop-left=p;break; case 2: stacktop-right=p;break; j+; ch=strj; void printree(BTree *b) if(b!=NULL) printf(%c,b-data); if(b-left!=NULL|b-right!=NULL) printf(%c,(); printree(b-left); if(b-right!=NULL) printf(%c,); printree(b-right); printf(%c,); int leafs(BTree *b)int num1,num2;if(b=NULL) return 0;else if(b-left=NULL&b-right=NULL) return 1;else num1=leafs(b-left); num2=leafs(b-right); return(num1+num2); main() BTree *b; creatree(b,(a(b(c),d(e(,f),g); printf(b xiangqian kuohao biaoshifa:);printree(b); printf(n); printf(yezhi jiedian geshu:,leafs(b),n); 三、 调试分