1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

给定后序和中序，求层次遍历。

好好背模板。

#include<stdio.h>#include<stdlib.h>#include<queue>using namespace std;int post[35],in[35];typedef struct Node *node;struct Node{int x;node left;node right;};node build(int s1,int e1,int s2,int e2){node T=NULL;if(!T){T=(node)malloc(sizeof(struct Node));T->left=T->right=NULL;T->x=post[e1];}int root,i;for(i=s2;i<=e2;i++){if(in[i]==post[e1]){root=i;break;}}if(root!=s2){T->left=build(s1,root-s2+s1-1,s2,root-1);}if(root!=e2){T->right=build(root-s2+s1,e1-1,root+1,e2);}return T;}int main(){int n,i;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&post[i]);}for(i=0;i<n;i++){scanf("%d",&in[i]);}node T=NULL;T=build(0,n-1,0,n-1);int flag=0;queue<node>q;q.push(T);while(!q.empty()){node head=q.front();q.pop();if(flag==0){printf("%d",head->x);flag=1;}else{printf(" %d",head->x);}if(head->left){q.push(head->left);}if(head->right){q.push(head->right);}} }