求助PAT:ZigZagging on a Tree

2019-10-12 01:19:35 303 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

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PAT真题练习（甲级）1127 ZigZagging on a Tree (30 分) PAT真题练习（甲级）1127 ZigZagging on a Tree (30 分) Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder t... 阅读详情

#include <stdio.h>



typedef int ElementType;



typedef struct TreeNode *Tree;

struct TreeNode

{

	ElementType Element;

	Tree Left;

	Tree Right;

	int Flag;

};



#define MAXN 30



Tree CreateNode(ElementType X);

Tree CreateTree(ElementType *InOrder, ElementType *PostOrder, int N, int cnt);

void LevelOrder(Tree T);



int Num=0;



int main()

{

	int N;

	ElementType InOrder[MAXN], PostOrder[MAXN];

	Tree BinaryTree;

	int i;



	scanf("%d", &N);



	Num = N;



	for (i = 0; i < N; i++)

		scanf("%d", &InOrder[i]);



	for (i = 0; i < N; i++)

		scanf("%d", &PostOrder[i]);



	BinaryTree = CreateTree(InOrder, PostOrder, N, 1);



	LevelOrder(BinaryTree);



	system("pause");

	return 0;

}



Tree CreateNode(ElementType X)

{

	Tree T;



	T = (Tree)malloc(sizeof(struct TreeNode));



	T->Element = X;



	T->Left = NULL;

	T->Right = NULL;



	return T;

}



Tree CreateTree(ElementType *InOrder, ElementType *PostOrder, int N, int cnt)

{

	Tree T;

	ElementType InTreeLeft[MAXN], InTreeRight[MAXN], PostTreeLeft[MAXN], PostTreeRight[MAXN];

	int i, j, k = 0;

	int N_Left, N_Right;



	if (N == 0)

		return NULL;

	else

	{

		T = CreateNode(PostOrder[N - 1]);

		T->Flag = cnt;

		if (N != 1)

		{

			for (i = 0; InOrder[i] != PostOrder[N - 1]; i++)

				InTreeLeft[i] = InOrder[i];



			for (j = i + 1; j < N; j++)

				InTreeRight[k++] = InOrder[j];



			N_Left = i;

			N_Right = k;



			for (i = 0; PostOrder[i] != InTreeRight[0]; i++)

				PostTreeLeft[i] = PostOrder[i];



			k = 0;

			for (j = i; j < N - 1; j++)

				PostTreeRight[k++] = PostOrder[j];



			cnt++;

			T->Left = CreateTree(InTreeLeft, PostTreeLeft, N_Left, cnt);

			T->Right = CreateTree(InTreeRight, PostTreeRight, N_Right, cnt);

		}



		return T;

	}

}



void LevelOrder(Tree T)

{

	int result[30][16] = { 0 };

	Tree TreeNow;

	Tree Queue[30];

	int Front=0, Rear=0;

	int i[30] = { 0 };

	int j, k, l = 0;

	int cnt = 1;



	Queue[Rear++] = T;

	//当队头和队尾相等时，表示队列为空

	while (Front < Rear)

	{

		//队头结点出队

		TreeNow = Queue[Front++];



		result[TreeNow->Flag - 1][i[TreeNow->Flag - 1]] = TreeNow->Element;

		i[TreeNow->Flag - 1]++;



		//将队头结点的左右孩子依次入队

		if (TreeNow->Left != NULL) 

		{

			Queue[Rear++] = TreeNow->Left;

		}

		if (TreeNow->Right != NULL) 

		{

			Queue[Rear++] = TreeNow->Right;

		}

	}



	for (j = 0; cnt < Num; j++)

	{

		if (j % 2 != 0)

		{

			for (k = 0; result[j][k]!=0; k++)

			{

				printf("%d", result[j][k]);

				if (cnt < Num)

					printf(" ");

				cnt++;

			}

		}

		else

		{

			while (result[j][l] != 0)

				l++;



			for (k = l - 1; k >= 0; k--)

			{

				printf("%d", result[j][k]);

				if (cnt < Num)

					printf(" ");

				cnt++;

			}

		}

	}

}