获取二叉树叶子节点个数

//获取二叉树叶子节点个数

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

//用二叉链表存储方式建树(完全二叉树)
typedef struct BitTree {
	int data;
	struct BitTree* LChild; //左子树
	struct BitTree* RChild; //右子树
}bittree;

//创建二叉树
bittree* createBitTree(bittree* BT) {
	int num = 0;
	scanf("%d", &num);
	if (num != -1) {  //输入-1代表结束
		BT = (bittree*)malloc(sizeof(bittree));
		BT->data = num;
		printf("输入%d的左结点值：", BT->data);
		BT->LChild = createBitTree(BT->LChild);
		printf("输入%d的右结点值：", BT->data);
		BT->RChild = createBitTree(BT->RChild);
	}
	else {
		BT = NULL; //输入-1，结点为NULL
	}
	return BT;
}
//先序遍历
void PrePrint(bittree* BT) {
	if (BT != NULL) {
		printf("%d ", BT->data);
		PrePrint(BT->LChild);
		PrePrint(BT->RChild);
	}
	else { //结点为NULL，返回上一层
		return;
	}
}


//获取二叉树叶子节点个数
int getLeaf(bittree* BT) {
	int count;
	if (BT == NULL) {
		count = 0;
	}
	else if (BT->LChild == NULL && BT->RChild == NULL) {
		count = 1;
	}		
	else {
		count = getLeaf(BT->LChild) + getLeaf(BT->RChild);
	}
	return count;
}

void main() {
	bittree* myBT = NULL;
	myBT = createBitTree(myBT);
	printf("先序遍历二叉树：
");
	PrePrint(myBT);
	printf("
");
	printf("二叉树的叶子节点个数是：%d", getLeaf(myBT));
}