堆与二叉树（下）

2023-12-24 23:27:57

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;
}

文章来源:https://blog.csdn.net/2301_79789645/article/details/135187641
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