第一次用了建树的方法，比较繁琐，《数据结构》课程中又介绍了不用建树的思路，根据先序和中序直接得到后序遍历的结果，代码如下：
1 void solve( int preL, int inL, int postL, int n) //当前处理树的三种遍历的第一个元素和树的规模 2 { 3 if(n==0) return; //边界条件 4 if(n==1) { 5 post[postL] = pre[preL]; 6 return; 7 } 8 root = pre[preL]; //当前树的根为先序的第一个元素 9 post[postL+n-1] = root; //将根节点的值填到后序遍历的数组中的最后一个 10 for(i = 0; i < n; i++) 11 if( in[inL+i] ==root) break; //找到根节点在中序遍历中的位置 12 L = i; R = n - L - 1; //计算左子树的规模和右子树的规模 13 solve(preL+1,inL,postL,L); //递归 14 solve(preL+L+1,intL+L+1,postL+L,R); 15 }
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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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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
AC代码
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 5 typedef struct BTNode{ 6 7 int Data; 8 struct BTNode *Right; 9 struct BTNode *Left; 10 11 }*BT; 12 void InputData(int preorder[],int inorder[]); 13 BT CreatBinaryTree(int preorder[],int inorder[],int N,int *pre_index); 14 void PostTravelofBT(BT T,int *k); 15 int main(){ 16 17 int N; 18 BT T; 19 int k = 1; 20 int pre_index = 0; 21 scanf("%d ",&N); 22 int preorder[N],inorder[N]; 23 InputData(preorder,inorder); 24 T = CreatBinaryTree(preorder,inorder,N,&pre_index); 25 PostTravelofBT(T,&k); 26 27 return 0; 28 } 29 30 void InputData(int preorder[],int inorder[]){ 31 int temp[30]; 32 int top = -1; 33 char inputstring[20]; 34 int indexforpre = 0; 35 int indexforin = 0; 36 while(gets(inputstring)){ 37 if(strcmp(inputstring,"Pop")==0){ 38 inorder[indexforin] = temp[top]; 39 indexforin ++; 40 top --; 41 }else{ 42 int data_tem; 43 sscanf(inputstring,"%*c%*c%*c%*c %d",&data_tem); 44 preorder[indexforpre] = data_tem; 45 indexforpre++; 46 top++; 47 temp[top] = data_tem; 48 } 49 50 } 51 } 52 53 BT CreatBinaryTree(int preorder[],int inorder[],int N,int *Ppre_index){ 54 BT T; 55 int i; 56 T = (BT)malloc(sizeof(struct BTNode)); 57 T->Data = preorder[*Ppre_index]; 58 (*Ppre_index)++; 59 T->Left = NULL; 60 T->Right = NULL; 61 for( i = 0; i <N; i++){ 62 if(inorder[i]== T->Data){ 63 inorder[i] = -1; 64 if(i!=0&&inorder[i-1]!=-1){ 65 T->Left = CreatBinaryTree(preorder,inorder,N,Ppre_index); 66 } 67 if(i!=N-1&&inorder[i+1]!=-1){ 68 T->Right = CreatBinaryTree(preorder,inorder,N,Ppre_index); 69 } 70 } 71 } 72 73 return T; 74 } 75 76 77 void PostTravelofBT(BT T,int *k){ 78 if(T){ 79 PostTravelofBT(T->Left,k); 80 PostTravelofBT(T->Right,k); 81 if(*k ==1){ 82 printf("%d",T->Data); 83 *k = 0; 84 }else{ 85 printf(" %d",T->Data); 86 } 87 } 88 return ; 89 }
