FDS编程练习 · 其一

日常练手用，熟悉一下几种结构的代码实现。

1.通过给定中序遍历和后序遍历的数列重建二叉树

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

typedef struct Node *Tree;
struct Node
{
	int data;
	Tree left, right;
};
/*定义树*/

Tree buildTree (int inorder[], int postorder[], int n);
/*建树函数*/

void preorderTraversal (Tree theTree);
/*先序遍历函数（用于验证）*/

int main ()
{
	int n, i;
	scanf ("%d", &n);
	/*输入含 n 个数的数列*/
	
	int inorder[n], postorder[n];
	for (i = 0; i < n; i++)
		scanf ("%d", &inorder[i]);
	for (i = 0; i < n; i++)
		scanf ("%d", &postorder[i]);
	/*输入中序和后序数列*/
	
	Tree theTree = buildTree (inorder, postorder, n);
	/*建树*/
	
	preorderTraversal (theTree);
	/*遍历重建的树*/
	
	return 0; 
}

Tree buildTree (int inorder[], int postorder[], int n)
{
	Tree T = (Tree)malloc(sizeof (struct Node));
	T->data = postorder[n - 1];
	/*后序遍历的最后一个数即根结点数据*/
	
	if (n == 1)
	{
		T->left = NULL;
		T->right = NULL;
		return T;
	}
	/*若只剩一个数，则直接返回（别忘了令左右子树为 NULL ）*/
	
	int i;
	for (i = 0; i < n; i++)
	{
		if (inorder[i] == postorder[n - 1])
			break;
	}
	if (i == 0)
	{
		T->left = NULL;
		T->right = buildTree (&inorder[1], postorder, n - 1);
	}
	/*中序遍历的第一个数是树的最左端，故该 T 无左子树，从中序遍历第二项开始继续建树*/
	
	else if (i == n-1)
	{
		T->right = NULL;
		T->left = buildTree (inorder, postorder, n - 1);
	}
	/*中序遍历的最后一个是树的最右端，故该 T 无右子树，再次遍历至倒数第二项为止*/
	
	else//此时 T 既有左子树又有右子树 
	{
		T->left = buildTree (inorder, postorder, i);
		/*用前 i 个数建左子树*/
		 
		T->right = buildTree (&inorder[i + 1], &postorder[i], n - 1 - i);
		/*用后 n-i 个数建右子树（ n-1-i 即 n-(1+i) ）*/ 
	}
	 
	return T;
} 

void preorderTraversal (Tree theTree)
{
	Tree T = theTree;
	if (T != NULL)
	{
		printf ("%d ", T->data);
		preorderTraversal (T->left);
		preorderTraversal (T->right);
	}
}

2.通过给定中序遍历和先序遍历的数列重建二叉树

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

typedef struct Node *Tree;
struct Node
{
	int data;
	Tree left, right;
};
/*定义树*/

Tree buildTree (int inorder[], int preorder[], int n);
/*建树函数*/

void postorderTraversal (Tree theTree);
/*后序遍历函数（用于验证）*/

int main ()
{
	int n, i;
	scanf ("%d", &n);
	/*输入含 n 个数的数列*/
	
	int inorder[n], preorder[n];
	for (i = 0; i < n; i++)
		scanf ("%d", &inorder[i]);
	for (i = 0; i < n; i++)
		scanf ("%d", &preorder[i]);
	/*输入中序和后序数列*/
	
	Tree theTree = buildTree (inorder, preorder, n);
	/*建树*/
	
	postorderTraversal (theTree);
	/*遍历重建的树*/
	
	return 0; 
}

Tree buildTree (int inorder[], int preorder[], int n)
{
	Tree T = (Tree)malloc(sizeof (struct Node));
	T->data = preorder[0];
	/*先序遍历的第一个数即根结点数据*/
	
	if (n == 1)
	{
		T->left = NULL;
		T->right = NULL;
		return T;
	}
	/*只剩一个数时直接返回*/
	
	int i;
	for (i = 0; i < n; i++)
	{
		if (inorder[i] == preorder[0])
			break;
	}
	if (i == 0)
	{
		T->left == NULL;
		T->right = buildTree (&inorder[1], &preorder[1], n - 1);
	}
	else if (i == n-1)
	{
		T->right = NULL;
		T->left = buildTree (inorder, &preorder[1], n - 1);
	} 
	else
	{
		T->left = buildTree (inorder, &preorder[1], i);
		T->right = buildTree (&inorder[i + 1], &preorder[i + 1], n - i - 1);
	} 
	/*原理类比上个代码*/
	 
	return T;
} 

void postorderTraversal (Tree theTree)
{
	Tree T = theTree;
	if (T != NULL)
	{
		postorderTraversal (T->left);
		postorderTraversal (T->right);
		printf ("%d ", T->data);
	}
}