03-树3 Tree Traversals Again(浙大数据结构PTA习题）

03-树3 Tree Traversals Again        分数 25        作者 陈越

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 (≤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:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

代码长度限制：16 KB        时间限制：400 ms        内存限制：64 MB

题目解析：

本质是根据一棵树的先序遍历结果和中序遍历结果，求解这棵树的后序遍历结果。

在树的遍历中，知中序和先序可求唯一后序；知中序和后序可求唯一先序；知先序和后序不可求唯一中序。

关键点1：根据输入获得先序遍历结果和中序遍历结果

关键点2：根据先序和中序遍历结果输出后序遍历结果

下面将演示建树和不建树两种情况

代码展示：

不建树的情况

// 根据先序遍历与中序遍历求后序遍历# include<stdio.h>
# include<string.h>
# define MAXSIZE 5void Solve(int preL, int inL, int postL, int n, int pre[], int in[], int post[]);int main(){// 一共有多少个结点 int N;scanf("%d",&N);// 创建一个堆栈int Stack[N];int Rear = -1;// 分别创建前、中、后序遍历的结果数组int Pre[N],In[N],Post[N];int PreRear = -1;// count作为弹出元素的计数int num, count = 0;while(count!=N){char handle[MAXSIZE];scanf("%s",handle);if(!strcmp(handle,"Push")){// 接受一个数字并压入栈scanf("%d",&num);Stack[++Rear] = num; // 前序遍历数组记录Pre[++PreRear] = num; }else{// 出栈并在中序遍历中进行记录 num = Stack[Rear--];In[count++] = num;}} // 有了前序与中序遍历结果，接下来通过递归函数求后序遍历结果 Solve(0,0,0,N,Pre,In,Post);int i;for(i=0;i<N;i++){if(i!=N-1)printf("%d ",Post[i]);else printf("%d",Post[i]);}
} void Solve(int preL, int inL, int postL, int n, int pre[], int in[], int post[]){if(n==0)return;if(n==1){post[postL] = pre[preL];return;}// 先序的第一个放在给定后序的最后一个 int root = pre[preL];post[postL+n-1] = root;// 找出左子树的范围 int i;for(i=0;i<n;i++){if(in[inL+i]==root)break;}// 左子树要处理的的结点数 int L = i;// 右子树要处理的结点数 int R = n-L-1;// 递归处理 Solve(preL+1, inL, postL,L,pre,in,post);Solve(preL+L+1,inL+L+1,postL+L,R,pre,in,post);return; 
}

建树的情况：

# include<stdio.h>
# include<stdlib.h>
# include<string.h>
# define MAXNODE 30
# define MAXLENGTH 5typedef struct TreeNode* Tree;
struct TreeNode{int data;Tree left;Tree right;
}; void PostOrder(Tree root,int data);
void CreateTree(int left1,int right1,int left2,int right2,int Pre[],int In[], Tree root);int main(){// 创建一个堆栈int Stack[MAXNODE];int SHead = -1; // 接收结点个数 int N;scanf("%d",&N);getchar();// 分别创建一个用来记录先序和中序遍历结果的数组int PreArray[N],InArray[N];int PreHead=-1, InHead=-1;// 接收操作输入，因为有出入栈，因此共有2*N次操作 int i,num;char* str = (char*)malloc(sizeof(char)*MAXLENGTH);for(i=0;i<2*N;i++){scanf("%s",str);if(strcmp(str,"Push")==0){// 入栈scanf("%d",&num);getchar();Stack[++SHead] = num;// 记录先序PreArray[++PreHead] = num; }else{// 出栈num = Stack[SHead--];// 记录中序InArray[++InHead] = num; }}// 创建根结点Tree root = (Tree)malloc(sizeof(struct TreeNode));root->data = PreArray[0];root->left = NULL;root->right = NULL;// 构建树CreateTree(0,N-1,0,N-1,PreArray,InArray,root);// 后序输出遍历结果，为了格式化输出因此传入根结点的值作为判读依据 PostOrder(root,root->data); return 0; 
}// 递归后序遍历树
void PostOrder(Tree root,int data){if(root){PostOrder(root->left,data);PostOrder(root->right,data);if(root->data == data)printf("%d",root->data);else printf("%d ",root->data);}return;
} // 根据先序和后序构建一棵树；left1与right1表示先序序列的范围；left2与right2表示后序序列的范围
void CreateTree(int left1,int right1,int left2,int right2,int Pre[],int In[], Tree root){// 首先找到根结点在中序中的位置int i,index;for(i=left2;i<=right2;i++){if(root->data == In[i]){index = i;break;}} // 如果根结点有左孩子 if(index!=left2){// 创建一个左孩子结点Tree left_son = (Tree)malloc(sizeof(struct TreeNode));left_son->data = Pre[left1+1];left_son->left = NULL;left_son->right = NULL;root->left = left_son;// 递归构建这个左孩子结点CreateTree(left1+1,index-left2+left1,left2,index-1,Pre,In,left_son); }// 如果根结点有右孩子if(index!=right2){Tree right_son = (Tree)malloc(sizeof(struct TreeNode));right_son->data = Pre[index-left2+left1+1];right_son->left = NULL;right_son->right = NULL;root->right = right_son;// 递归构建这个右孩子结点CreateTree(index-left2+left1+1,right1,index+1,right2,Pre,In,right_son); } return;
} 

运行结果：

两种情况均可通过测试点