C二叉树基础

c二叉树基础：c二叉树又该怎么实现呢？希望下面的文章对大家有所帮助。

#include
#include"linkqueue.h"
//#include"linkstack.h"

typedef char data_t;
typedef struct btnode {
	data_t data;
	struct btnode* lchild;
	struct btnode* rchild;
}btree_t;

 void* assertx(void* p)
{
	if(p == null) {
		printf("argument error\n");
		return null;
	}
	
}

/**
 *	@data: 节点数据指针
 *	@num:  节点位置编号从根1开始
 *	@len:  待存入树节点中的数据的长度
 */
btree_t* create_btree(data_t* data, int num, int len)
{
	btree_t *root;
	// 1.
	root = (btree_t*)malloc(sizeof(btree_t));
	if(root == null) {
		printf("allocate memory error\n");
		return null;
	}
	// 2.
	root->data = data[num];
	root->lchild = null;
	root->rchild = null;
	// 3. 这里利用满二叉树的特性
	if(2*num <= len) {
		root->lchild = create_btree(data, 2*num, len);

	}
	if(2*num+1 <= len) {
		root->rchild = create_btree(data, 2*num+1, len);
	}

	return root;
}
// 层次遍历
int btree_travers_layer(btree_t* root)
{
	assertx(root);
	btree_t* temp;
	linkqueue* queue = create_empty_linkqueue();

	// 1. root 入队列
	enter_linkqueue(queue, root);
	// 2. root 出队列，每出一个，拿到其孩子节点放入队尾
	// 直到队列为空，完成遍历
	while(!is_empty_linkqueue(queue)) {
		temp = delete_linkqueue(queue);
		printf("[%c] ", temp->data);
		if(temp->lchild != null) 
			enter_linkqueue(queue, temp->lchild);
		if(temp->rchild != null)
			enter_linkqueue(queue, temp->rchild);
	}
	printf("\n");
}
// 前序遍历
int btree_travers_preorder(btree_t* root)
{	
	if(root == null) {
		return -1;
	}
	printf("[%c] ", root->data);
	if(root->lchild != null) {
		btree_travers_preorder(root->lchild);
	}
	if(root->rchild != null) {
		btree_travers_preorder(root->rchild);
	}

	return 0;

}


// 中序遍历
int btree_travers_midorder(btree_t* root)
{	
	if(root == null) {
		return -1;
	}
	if(root->lchild != null) {
		btree_travers_midorder(root->lchild);
	}
	printf("[%c] ", root->data);
	if(root->rchild != null) {
		btree_travers_midorder(root->rchild);
	}

	return 0;

}
// 后序遍历
int btree_travers_postorder(btree_t* root)
{	
	if(root == null) {
		return -1;
	}
	if(root->lchild != null) {
		btree_travers_postorder(root->lchild);
	}
	if(root->rchild != null) {
		btree_travers_postorder(root->rchild);
	}
	printf("[%c] ", root->data);

	return 0;

}
/*****************************************************
 * 计算树的高度
 *	思路: 
 *	    用两个队列a，b。第一层节点入a队列，弹出a
 *	队列中的元素所有元素，每出一个，判读其左右孩子
 *	若果存在孩子，则孩子入b队列。层数累加。然后弹出
 *	所有b的元素，同样每出一个，判断左右孩子，存在则
 *	入a队列，层数累加。如此直到队列为空。
 *****************************************************/
unsigned int count_btree_height(btree_t* root)
{
	unsigned int count = 0;
	btree_t* temp;
	linkqueue*	aq = create_empty_linkqueue();
	linkqueue*	bq = create_empty_linkqueue();
	linkqueue*  	tq;

	if(root == null) {
		return 0;
	}

	enter_linkqueue(aq, root);
	while(!is_empty_linkqueue(aq)) {		// 判断也就是上一层的孩子构成的队列是否为空
		while(!is_empty_linkqueue(aq)) {	// 出队直到整个队列元素都被处理
			temp = delete_linkqueue(aq);	// 
			if(temp->lchild != null) {
				enter_linkqueue(bq, temp->lchild);
			}
			if(temp->rchild != null) {
				enter_linkqueue(bq, temp->rchild);
			}
		}
		count++;
		tq = aq;		// 为了简化代码： 交换 aq和bq，就是下次操作之前层的孩子队列
		aq = bq;
		bq = tq;
	}
	
	return count;
}



int main()
{
	char ca[] = {  0, 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j',  'k', 'l', 'm', 'n'};
	btree_t* tree = create_btree(ca, 1, sizeof(ca)/sizeof(ca[0]) - 1);

	printf("layer travers: ");
	btree_travers_layer(tree);

	printf("pre   travers: ");
	btree_travers_preorder(tree);

	printf("\n");

	printf("mid   travers: ");
	btree_travers_midorder(tree);

	printf("\n");

	printf("post  travers: ");
	btree_travers_postorder(tree);
	printf("\n");
	printf("btree height: %d\n", count_btree_height(tree));

	return 0;

}

这里使用了队列和递归的思维解决问题。

gcc输出：

layer travers: [a] [b] [c] [d] [e] [f] [g] [h] [i] [j] [k] [l] [m] [n]
pre  travers: [a] [b] [d] [h] [i] [e] [j] [k] [c] [f] [l] [m] [g] [n]
mid  travers: [h] [d] [i] [b] [j] [e] [k] [a] [l] [f] [m] [c] [n] [g]
post travers: [h] [i] [d] [j] [k] [e] [b] [l] [m] [f] [n] [g] [c] [a]
btree height: 4