【PAT】1020. Tree Traversals

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

32000 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

题意：根据后序遍历、中序遍历结果，生成 层次遍历。

代码如下：

#include <iostream>#include <fstream>using namespace std;ifstream fin("in.txt");#define cin fin#define MAXLEN 31int postOrder[MAXLEN];int inOrder[MAXLEN];int store[MAXLEN][17]={0};int findIndex(int key,int inBegin,int inEnd)//在中序遍历中寻找父节点{for(int i=inBegin;i<=inEnd;i++){if(inOrder[i]==key)return i;}return -1;}void levelOrder(int begin,int end,int inBegin,int inEnd,int level){if(begin>end)return;store[level][0]++;store[level][store[level][0]] = postOrder[end];if(begin==end)return;int inIndex=findIndex(postOrder[end],inBegin,inEnd);int postIndex = begin + inIndex - inBegin;//根据inIndex左子节点的数目来计算后序遍历的中间点levelOrder(begin,postIndex-1,inBegin,inIndex-1,level+1);levelOrder(postIndex,end-1,inIndex+1,inEnd,level+1);}int main(){int n;cin>>n;int i;for(i=0;i<n;i++)cin>>postOrder[i];for(i=0;i<n;i++)cin>>inOrder[i];levelOrder(0,n-1,0,n-1,0);cout<<store[0][1];int le = 1;int num = store[le][0];while(num){for(i=0;i<num;i++){cout<<' '<<store[le][i+1];}le++;num = store[le][0];}cout<<endl;system("PAUSE");return 0;}