1119. Pre- and Post-order Traversals (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Special

作者

CHEN, Yue

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 traversal sequences, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique.

Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first printf in a line "Yes" if the tree is unique, or "No" if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input 1:

71 2 3 4 6 7 52 6 7 4 5 3 1

Sample Output 1:

Yes2 1 6 4 7 3 5

Sample Input 2:

41 2 3 42 4 3 1

Sample Output 2:

No2 1 3 4

给出前序和后序，问中序是否唯一。唯一则输出，不唯一输出可能的中序

参考算法笔记：

如果前序的第二个点和后序的倒数第二点相同，则根节点只有左孩子或右孩子。树就不唯一。

注意：

1.当s1==e1时，也就是一个结点的时候，直接返回。

2.最后居然还有一个换行。。。

#include<stdio.h>#include<stdlib.h>typedef struct Node *node;struct Node{int x;node left;node right;};int pre[35],post[35];int flag=1,printflag=1;node build(int s1,int e1,int s2,int e2){node T=(node)malloc(sizeof(struct Node));T->x=pre[s1];T->left=T->right=NULL;if(s1==e1){//!!!return T;}int index,i;for(i=s2;i<=e2-1;i++){if(pre[s1+1]==post[i]){index=i;break;}}if(index==e2-1){//只有一个孩子结点 ，挂在左边右边都可以 flag=0;T->left=build(s1+1,e1,s2,e2-1);}else{T->left=build(s1+1,index-s2+s1+1,s2,index);T->right=build(index-s2+s1+2,e1,index+1,e2-1);}return T;}void inorder(node T){if(T){inorder(T->left);if(printflag==1){printf("%d",T->x);printflag=0;}else{printf(" %d",T->x);}inorder(T->right);}} int main(){int n,i;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&pre[i]);}for(i=0;i<n;i++){scanf("%d",&post[i]);}node T=build(0,n-1,0,n-1);if(flag==1){printf("Yes\n");}else{printf("No\n");}inorder(T);printf("\n");//晕死。。。。 }