【C语言数据结构】堆与二叉树（下）

admin • 2024-04-08 18:55 • C/C++

接着上次的，这里主要介绍的是堆排序，二叉树的遍历，以及之前讲题时答应过的简单二叉树问题求解

堆排序

给一组数据，升序（降序）排列

思路

思考：如果排列升序，我们应该建什么堆？

首先，如果排升序，数列最后一个数是最大数，我们的思路是通过向上调整或者向下调整，数组存放的第一个数不是最小值（小堆）就是最大值（大堆），此时我们将最后一个数与第一个数交换，使得最大值放在最后，此时再使用向上调整或者向下调整，得到第二大的数，重复上述动作，很明显，我们需要的第一个数是最大值，因此我们需要建大堆

反之，排降序，建立小堆

代码

#include<stdio.h>
void downAdjust(int* pa, int parent, int n)
{
	int child = parent * 2 + 1;
	while (child < n)
	{
		if (child + 1 < n && pa[child] > pa[child + 1])
		{
			child++;
		}
		if (pa[parent] > pa[child])
		{
			swap(&pa[parent], &pa[child]);
		}
		else
		{
			break;
		}
		parent = child;
		child = parent * 2 + 1;
	}
}
int main()
{
	int arr[] = { 1,3,2,5,7,4,7,4,2,5,6,8};
	int n = sizeof(arr) / sizeof(arr[0]);
	  for (int i = (n - 1 - 1) / 2; i >= 0; i--)
	  {
		downAdjust(arr, i, n);
	  }
		for (int i = n; i > 0; )
		{
			swap(&arr[0], &arr[i - 1]);
			downAdjust(arr, 0, --i);
		}
		for (int i = 0; i < n; i++)
		{
			printf("%d ", arr[i]);
		}

	return 0;
}

topK算法

在一组数据中，选出k个最大（最小）的数

思路

如果我们选择k个最大的数，假设数组的前k个数就是最大的数，这 k个数建立小堆，带一个数与后面的从第 k + 1个数开始，进行比较，如果比第一个数的就换下来，然后向下调整，直到每个所有数都比较完了

代码

void downAdjust(int* pa, int parent, int n)
{
	int child = parent * 2 + 1;
	while (child < n)
	{
		if (child + 1 < n && pa[child] > pa[child + 1])
		{
			child++;
		}
		if (pa[parent] > pa[child])
		{
			swap(&pa[parent], &pa[child]);
		}
		else
		{
			break;
		}
		parent = child;
		child = parent * 2 + 1;
	}
}
#include<stdio.h>
int main()
{
	int arr[] = { 1,6,10,3,5,8,46,23,6,25,3,40 };
	int n = sizeof(arr) / sizeof(arr[0]);
	int k = 0;
	scanf("%d", &k);
	for (int i = (k - 1 - 1) / 2; i >= 0; i--)
	{
		downAdjust(arr, i, n);
	}
	for (int i = k; i < n; i++)
	{
		if (arr[i] > arr[0])
		{
			swap(&arr[i], &arr[0]);
			downAdjust(arr, 0, k);
		}
	}
	for (int i = 0; i < k; i++)
	{
		printf("%d ", arr[i]);
	}
	return 0;
}

五. 二叉树的实现

1. 链接结构搭建二叉树

代码

typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL *creatnode(TLType x)
{
	TL*pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL *tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(3);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	return tree1;
}
#include<stdio.h>
int main()
{
	TL* p = NULL;
	p = CreatTree();
}

我们搭建了一个这样的树结构：

2. 二叉树的遍历

二叉树的遍历可以分三种：前序，中序，后序，层序

a. 前序遍历：（根，左子树，右子树）

举例

这棵树的前序遍历是怎样的？（包括空树，用N表示）

val1 val2 val4 N N val5 N N val3 val6 N N val7 N N

代码实现

#include<stdio.h>
#include<stdlib.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL *creatnode(TLType x)
{
	TL*pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL *tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	return tree1;
}
#include<stdio.h>
void PrevOrder(TL *root)
{
	if (root == NULL)
	{
		printf("N ");
		return;
	}
	printf("%d ", root->val);
	PrevOrder(root->left);
	PrevOrder(root->right);
}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	PrevOrder(p);
}

运行结果：

b. 中序遍历：（左子树，根，右子树）

举例

这棵树的中序遍历是怎样的？（包括空树，用N表示）

N val4 N val2 N val5 N val1 N val6 N val3 N val7 N

代码实现

#include<stdio.h>
#include<stdlib.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL *creatnode(TLType x)
{
	TL*pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL *tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	return tree1;
}
#include<stdio.h>
void InOder(TL* root)
{
	if (root == NULL)
	{
		printf("N ");
		return;
	}
	InOder(root->left);
	printf("%d ", root->val);
	InOder(root->right);
}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	InOder(p);
}

运行结果：

c. 后序遍历：（左子树，右子树，根）

举例

这棵树的后序遍历是怎样的？（包括空树，用N表示）

N N val4 N N val5 val2 N N val6 N N val7 val3 val1

代码实现

#include<stdio.h>
#include<stdlib.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	return tree1;
}
void PostOder(TL* root)
{
	if (root == NULL)
	{
		printf("N ");
		return;
	}
	PostOder(root->left);
	PostOder(root->right);
	printf("%d ", root->val);
}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	PostOder(p);
}

运行结果：

d. 层序遍历

一排排的遍历

画图举例

实现思路

这里我们借助队列（可以先进先出），开辟的数组里面存放根节点的地址（通过地址可以找到左右子树，否则如果存值是没有办法找到左右子树），打印完根节点的值，就释放，存入左右子树的节点

代码实现

实现的二叉树是这样的：

#include<stdio.h>
#include<stdlib.h>
#include<assert.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	return tree1;
}
typedef struct QueueNode
{
	struct QueueNode* next;
	TL* data;
}QNode;

typedef struct Queue
{
	QNode* head;
	QNode* tail;
	int size;
}Que;


void QueueInit(Que* pq)
{
	assert(pq);

	pq->head = pq->tail = NULL;
	pq->size = 0;
}

void QueueDestroy(Que* pq)
{
	assert(pq);

	QNode* cur = pq->head;
	while (cur)
	{
		QNode* next = cur->next;
		free(cur);
		cur = next;
	}

	pq->head = pq->tail = NULL;
	pq->size = 0;
}

void QueuePush(Que* pq, TL* x)
{
	assert(pq);

	QNode* newnode = (QNode*)malloc(sizeof(QNode));
	if (newnode == NULL)
	{
		perror("malloc fail");
		exit(-1);
	}

	newnode->data = x;
	newnode->next = NULL;

	if (pq->tail == NULL)
	{
		pq->head = pq->tail = newnode;
	}
	else
	{
		pq->tail->next = newnode;
		pq->tail = newnode;
	}

	pq->size++;
}

bool QueueEmpty(Que* pq)
{
	assert(pq);

	return pq->head == NULL;
}
void QueuePop(Que* pq)
{
	assert(pq);
	assert(!QueueEmpty(pq));

	if (pq->head->next == NULL)
	{
		free(pq->head);
		pq->head = pq->tail = NULL;
	}
	else
	{
		QNode* next = pq->head->next;
		free(pq->head);
		pq->head = next;
	}

	pq->size--;
}

TL* QueueFront(Que* pq)
{
	assert(pq);
	assert(!QueueEmpty(pq));

	return pq->head->data;
}




int QueueSize(Que* pq)
{
	assert(pq);

	return pq->size;
}
void leverOrder(TL* root, Que* pq)
{
	QueuePush(pq, root);
	while (!QueueEmpty(pq))
	{
		TL* pa = QueueFront(pq);
		printf("%d ", pa->val);
		QueuePop(pq);
		if (pa->left != NULL)
		{
			QueuePush(pq, pa->left);
		}
		if (pa->right != NULL)
		{
			QueuePush(pq, pa->right);
		}
	}

}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	Que q;
	QueueInit(&q);
	leverOrder(p, &q);
	return 0;
}

运行结果：

3. 简单二叉树经典问题求解

a. 求二叉树的节点个数

思路

想要求二叉树的节点可以分成根节点 + 左子树 + 右子树

这里的遍历类似 前序遍历

代码

实现的树是这样的：

#include<stdio.h>
#include<stdlib.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	return tree1;
}
int TreeSize(TL* root)
{
	if (root == NULL)
	{
		return 0;
	}
	return 1 + TreeSize(root->left) + TreeSize(root->right);
}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	int size = TreeSize(p);
	printf("%d ", size);
	return 0;
}

b. 求树的高度

思路

求二叉树的高度，我们需要找到到那个最长的路径，这里采用分治的思想，如果为空树，返回 0 （空树高度为 0），调用左子树和右子树都会 + 1（+ 1可以理解成加上节点的高度），对比左子树和右子树，返回高度最大的那个

注：每求一次左右节点个数时，一定要保存，否则会有很大的时间浪费

代码

#include<stdio.h>
#include<stdlib.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	TL* tree8 = creatnode(8);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	tree4->left = tree8;
	return tree1;
}
int TreeHigh(TL* root)
{
	if (root == NULL)
	{
		return 0;
	}
	int Left = 1 + TreeHigh(root->left);
	int Right = 1 +  TreeHigh(root->right) ;
	return Left > Right ? Left : Right;
}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	int high = TreeHigh(p);
	printf("%d ", high);
	return 0;
}

c. 求根节点的个数

思路

判断是否是根节点的方法就是判断它的左右子树是否是空树，我们只需要遍历这棵树就行，但如果遍历时，根节点遇到空树这也是一种结束条件

代码

#include<stdio.h>
#include<stdlib.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	TL* tree8 = creatnode(8);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	tree4->left = tree8;
	return tree1;
}
int RootSize(TL* root)
{
	if (root == NULL)
	{
		return 0;
	}
	if (root->left == NULL && root->right == NULL)
	{
		return 1;
	}
	return RootSize(root->left) + RootSize(root->right);
}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	int root = RootSize(p);
	printf("%d ", root);
	return 0;
}

d. 求倒数第k排节点的个数

思路

这个可以是求树的高度的变形，将计数倒过来

代码

#include<stdio.h>
#include<stdlib.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	TL* tree8 = creatnode(8);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	tree4->left = tree8;
	return tree1;
}

int TreeHigh(TL* root)
{
	if (root == NULL)
	{
		return 0;
	}
	int Left = 1 + TreeHigh(root->left);
	int Right = 1 +  TreeHigh(root->right) ;
	return Left > Right ? Left : Right;
}
int RootKsize(TL* root,int n,int k)
{
	if (root == NULL)
	{
		return 0;
	}
	if (n == k)
	{
		return 1;
	}
	return RootKsize(root->left, n - 1, k) + RootKsize(root->right, n - 1, k);
}
int main()
{
	int k = 0;
	scanf("%d", &k);
	TL* p = NULL;
	p = CreatTree();
	int high = TreeHigh(p);
	int rootk = RootKsize(p, high, k);
	printf("%d ", rootk);
	return 0;
}

e. 判断是否是相同的树

思路

采用前序，先比较根节点是否相同，再比较左右子树是否相同

代码

#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree1()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	TL* tree8 = creatnode(8);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	tree4->left = tree8;
	return tree1;
}
TL* CreatTree2()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(3);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	return tree1;
}
bool IsSameTree(TL* root1,TL* root2)
{
	if (root1 == NULL && root2 == NULL)
	{
		return true;
	}
	if (root1 == NULL || root2 == NULL)
	{
		return false;
	}
	if (root1->val != root2->val)
	{
		return false;
	}
	return IsSameTree(root1->left, root2->left) && IsSameTree(root1->right, root2->right);
}
int main()
{
	TL* p = NULL;
	p = CreatTree1();
	TL* q = CreatTree2();
	printf("%d ", IsSameTree(p, q));
	return 0;
}

f. 找到某个值，返回节点的地址

思路

前序遍历完数组，如果对比左右子树，判断是否找到节点的地址

代码

#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
typedef int TLType;
typedef struct TreeList
{
	TLType val;
	struct TreeList* left;
	struct TreeList* right;
}TL;
TL* creatnode(TLType x)
{
	TL* pa = (TL*)malloc(sizeof(TL));
	if (pa == NULL)
	{
		perror("malloc");
		return;
	}
	TL* newnode = pa;
	newnode->left = newnode->right = NULL;
	newnode->val = x;
	return newnode;
}
TL* CreatTree()
{
	TL* tree1 = creatnode(1);
	TL* tree2 = creatnode(2);
	TL* tree3 = creatnode(2);
	TL* tree4 = creatnode(4);
	TL* tree5 = creatnode(5);
	TL* tree6 = creatnode(6);
	TL* tree7 = creatnode(7);
	TL* tree8 = creatnode(8);
	tree1->left = tree2;
	tree1->right = tree3;
	tree2->left = tree4;
	tree2->right = tree5;
	tree3->left = tree6;
	tree3->right = tree7;
	tree4->left = tree8;
	return tree1;
}
TL* FindRoot(TL* root,int m)
{
	if (root == NULL)
	{
		return NULL;
	}
	if (root->val == m)
	{
		return root;
	}
	TL* Left = FindRoot(root->left, m);
	TL* Right = FindRoot(root->right, m);
	if (Left == NULL && Right == NULL)
	{
		return NULL;
	}
	if (Left == NULL && Right != NULL)
	{
		return Right;
	}
	else 
	{
		return Left;
	}
	
}
int main()
{
	TL* p = NULL;
	p = CreatTree();
	int m = 0;
	scanf("%d", &m);
	TL *root = FindRoot(p,m);
	if (root == NULL)
	{
		printf("找不到n");
	}
	else
	{
		printf("%d ", root->val);
	}
	return 0;
}

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【C语言数据结构】堆与二叉树（下）

堆排序

topK算法

五. 二叉树的实现

1. 链接结构搭建二叉树

2. 二叉树的遍历

a. 前序遍历：（根，左子树，右子树）

b. 中序遍历：（左子树，根，右子树）

c. 后序遍历：（左子树，右子树，根）

d. 层序遍历

3. 简单二叉树经典问题求解

a. 求二叉树的节点个数

b. 求树的高度

c. 求根节点的个数

d. 求倒数第k排节点的个数

e. 判断是否是相同的树

f. 找到某个值，返回节点的地址

最新文章

分类

标签云

【C语言 数据结构】堆与二叉树（下）

堆排序

topK算法

五. 二叉树的实现

1. 链接结构搭建二叉树

2. 二叉树的遍历

a. 前序遍历：（根，左子树，右子树）

b. 中序遍历：（左子树，根，右子树）

c. 后序遍历：（左子树，右子树，根）

d. 层序遍历

3. 简单二叉树经典问题求解

a. 求二叉树的节点个数

b. 求树的高度

c. 求根节点的个数

d. 求倒数第k排节点的个数

e. 判断是否是相同的树

f. 找到某个值，返回节点的地址

最新文章

分类

标签云

【C语言数据结构】堆与二叉树（下）