03-树3 Tree Traversals Again

03-树3 Tree Traversals Again   (25分)

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer $N$ ($\le 30$) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to $N$). Then $2N$ lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1

题目思路：以先序遍历压栈，中序遍历pop，后序遍历输出

#include<stdio.h>#include<stdlib.h>#include<string.h>#define MaxTree 30#define ElementType int#define ERROR -1 #define BinTree int /*以先序遍历压栈，中序遍历pop，后序遍历输出*///struct TreeNode//{//ElementType Element;//BinTree Left;//BinTree Right;//}T1[MaxTree];//***************堆栈相关 *************************************** typedef int Position;struct SNode {    ElementType *Data; /* 存储元素的数组 */    Position Top;      /* 栈顶指针 */    int MaxSize;       /* 堆栈最大容量 */};typedef struct SNode *Stack; Stack CreateStack( int MaxSize ){    Stack S = (Stack)malloc(sizeof(struct SNode));    S->Data = (ElementType *)malloc(MaxSize * sizeof(ElementType));    S->Top = -1;    S->MaxSize = MaxSize;    return S;} bool IsFull( Stack S ){    return (S->Top == S->MaxSize-1);} bool Push( Stack S, ElementType X ){    if ( IsFull(S) ) {        printf("堆栈满");        return false;    }    else {        S->Data[++(S->Top)] = X;        return true;    }} bool IsEmpty( Stack S ){    return (S->Top == -1);} ElementType Pop( Stack S ){    if ( IsEmpty(S) ) {        printf("堆栈空");        return ERROR; /* ERROR是ElementType的特殊值，标志错误 */    }    else         return ( S->Data[(S->Top)--] );}ElementType GetTopElement(Stack S){return(S->Data[S->Top]);}//***************全局变量************************************************int preorder[MaxTree]={0}, inorder[MaxTree]={0}, postorder[MaxTree]={0};int pre_i = 0 , in_i = 0, post_i = 0; //***********************************************************************//通过中序遍历和前序遍历来得到后序遍历 void GetPostOrder(int pre_i,int in_i,int post_i,int N){if(N==0)//没有输入节点{return;} else if(N == 1){//只有根节点 postorder[post_i+N-1] = preorder[pre_i];return ;  }//后序遍历的最后一个节点和前序遍历的第一个节点为根节点int left,right,root;root = preorder[pre_i];postorder[post_i+N-1] = root;for(int i = 0; i < N; i++)//通过中序遍历来找到左子树和右子树 {if(inorder[in_i+i] == root){left = i;//left tree numnoderight = N - i - 1;//right tree num nodebreak;}}//遍历左子树和右子树GetPostOrder(pre_i+1, in_i,post_i,left);GetPostOrder(pre_i+left+1, in_i+left+1, post_i+left,right);  }//***********************************************************int main(){int data,N;char str[30];//freopen("test.txt", "r", stdin); scanf("%d\n",&N);Stack S1 = CreateStack(MaxTree);for(int i = 0; i < (2*N); i++ ){scanf("%s",str);if(!strcmp(str,"Push"))//strcmp比较两个字符串，相等输出0{scanf("%d",&data);preorder[pre_i++] = data;Push(S1,data);}else if(!strcmp(str,"Pop")){inorder[in_i++]= Pop(S1);}}GetPostOrder(0,0,0,N); int flag = 1;for(int j = 0; j < N; j++ ){if(flag){flag = 0;printf("%d",postorder[j]);}else{printf(" %d",postorder[j]);}}  return 0;}