C/C++: 二叉树叶子节点的个数 & 二叉树深度的求解

 其实这个比较简单，对于求深度的时候：用一个开关flag来标记是否要深度+1；

对于求叶子结点的个数的时候：用num1和num2来标记，如果有左子结点，那么在每个循环中num1置1；同理，若有右子结点，在每个循环中num2置1；综上，每个循环中只要num1和num2 都是0的时候，说明没有子结点，则可以判断是叶子结点，最终答案ans++即可

/**
*
* Althor: Hacker Hao
* Create: 2023.11.1
*
*/
#include 
using namespace std;
#define ElemType int
#define MAXSIZE 200
typedef struct BiTNode
{
	ElemType data;
	struct BiTNode* lchild, * rchild;
}BTNode;

BTNode* Create(int val)
{
	BTNode* node = (BTNode*)malloc(sizeof(BTNode));
	node->data = val;
	node->rchild = NULL;
	node->lchild = NULL;
	return node;
}

void PreOrder(BTNode* root)
{
	int cnt = 0, ans = 0;
	int flag = 0;
	BTNode* stack[MAXSIZE], * p;
	int top = 0;
	if (!root)
	{
		cout << "ERROR" << endl;
		exit(0);
	}

	stack[top] = root;
	while (top != -1)
	{
		int num1 = 0, num2 = 0;
		p = stack[top];
		top--;

		cout << p->data << " ";
		if (p->rchild)
		{
			stack[++top] = p->rchild;
			cnt++;
			flag = 1;
			num1 = 1;
		}

		if (p->lchild)
		{
			stack[++top] = p->lchild;
			num2 = 1;
			if (flag == 0)
			{
				cnt++;
			}
		}
		if (num1 == 0 && num2 == 0)
			ans++;
	}
	cout << endl;
	cout << "深度为:" << cnt << endl;
	cout << "叶子节点为:" << ans << endl;
}

int main()
{
	cin.tie(0), cout.tie(0);
	BTNode* A = Create(1);
	BTNode* B = Create(2);
	BTNode* C = Create(3);
	BTNode* D = Create(4);
	BTNode* E = Create(5);
	BTNode* F = Create(6);
	BTNode* G = Create(7);
	BTNode* H = Create(8);

	A->lchild = B;
	A->rchild = C;
	B->lchild = D;
	B->rchild = E;
	E->lchild = G;
	E->rchild = H;
	C->rchild = F;

	cout << "前序遍历结果是:";
	PreOrder(A);
	cout << endl;

	return 0;
}