# 递归计算二叉树的叶子节点个数

#include “stdio.h”
#include “malloc.h”

typedef struct BiTNode{
char data;   /*结点的数据域*/
struct BiTNode *lchild , *rchild;  /*指向左孩子和右孩子*/
} BiTNode , *BiTree;

/*创建一棵二叉树*/
void CreatBiTree(BiTree *T)
{
char c;
scanf(“%c”,&c);
if(c == ‘ ‘) *T = NULL;
else{
*T = (BiTNode * )malloc(sizeof(BiTNode));  /*创建根结点*/
(*T)->data = c;    /*向根结点中输入数据*/
CreatBiTree(&((*T)->lchild));  /*递归地创建左子树*/
CreatBiTree(&((*T)->rchild));  /*递归地创建右子树*/
}
}

int getLeavesConut(BiTree T) {
int leftLeavesCount;
int rightLeavesCount;
if (T == NULL) {
return 0;
}else if (T->lchild == NULL && T->rchild == NULL) {
return 1;
}else {
leftLeavesCount = getLeavesConut(T->lchild);
rightLeavesCount = getLeavesConut(T->rchild);
return leftLeavesCount + rightLeavesCount;
}
}
main()
{
BiTree T = NULL;                /*初始化T */
int count = 0;
printf(“Input some characters to create a binary tree \n”);
CreatBiTree(&T);                /*创建一棵二叉树*/
count =  getLeavesConut (T);    /*计算二叉树中叶子结点的个数 */
printf(“The number of leaves of BTree are %d\n”,count);
getchar();
getchar();
}

原文作者：张荣华_csdn
原文地址: https://blog.csdn.net/zrh_CSDN/article/details/81175136
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