Tree Traversals

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

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提交代码

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;const int maxn=50;struct node{int data;//数据域 node* lchild;//指向左子树根结点的指针 node* rchild; //指向右子树根结点的指针 }; int pre[maxn],in[maxn],post[maxn];int n;node* create(int postL,int postR,int inL,int inR){if(postL>postR){return NULL;}node* root=new node;root->data=post[postR];int k;for(k=inL;k<=inR;k++){if(in[k]==post[postR]){break;}}int numLeft=k-inL;root->lchild=create(postL,postL+numLeft-1,inL,k-1);root->rchild=create(postL+numLeft,postR-1,k+1,inR);return root;}int num=0;void BFS(node* root){queue<node*>q;q.push(root);while(!q.empty()){node* now=q.front();q.pop();printf("%d",now->data);num++;if(num<n) printf(" ");if(now->lchild!=NULL){q.push(now->lchild);} if(now->rchild!=NULL) {q.push(now->rchild);}}}int main(){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&post[i]);}for(int i=0;i<n;i++){scanf("%d",&in[i]);}node* root=create(0,n-1,0,n-1);BFS(root);return 0;}