统计C语言二叉树中叶子结点个数统计C语言二叉树中叶子结点个

树是一种非线性的数据结构，它是由n（n>=0）个有限结点组成一个具有层次关系的集合。把它叫做树是因
为它看起来像一棵倒挂的树，也就是说它是根朝上，而叶朝下的，下面我们就用简单小栗子来简单说明关于统计C语言二叉树中叶子结点个数的方法吧
简单小栗子：

#include#include typedef char ElemType; typedef struct BTNode{ElemType data; struct BTNode *left; struct BTNode *right; }BTNode,*BiTree; //创建二叉树void createBTNode(BiTree &BT){ElemType ch; scanf("%c",&ch); if(ch==' ')BT=NULL; else{BT = (BTNode*)malloc(sizeof(BTNode)); BT->data= https://www.it610.com/article/ch; createBTNode(BT->left); createBTNode(BT->right); }} //先序遍历二叉树void printDLR(BiTree BT){if(BT){printf("%c ",BT->data); printDLR(BT->left); printDLR(BT->right); }} //统计二叉树结点个数void countLeaves(BiTree BT,int &count){if(BT){if(BT->left==NULL && BT->right==NULL)count++; else{countLeaves(BT->left,count); countLeaves(BT->right,count); }}} void main(){BTNode *BT; int count=0; createBTNode(BT); printf("先序遍历："); printDLR(BT); printf("\n"); countLeaves(BT,count); printf("二叉树结点的个数：%d\n",count); }

按照先序遍历的方式来输入二叉树结点，若孩子结点为空，则输入空格。
输入：