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:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

分析:考察树的建立和遍历。

课参考《编程之美》3.9

#include<iostream>
#include<map>
#include<vector>
#include<queue>
using namespace std; struct Node{
Node *left;
Node *right;
int value;
Node():left(NULL),right(NULL){}
}; void Rebuild(int * PostOrder, int * InOrder, int len, Node* &root){
//判断何时结束递归
if(PostOrder == NULL || InOrder == NULL)
{
root = NULL;
return ;
}
if(root == NULL) root = new Node;
root->value = *(PostOrder + len - 1);
root->left = NULL;
root->right = NULL;
if(len == 1)
return; int count = 0;
int *temp = InOrder;
while(*temp != *(PostOrder + len -1))
{
count ++;
temp++;
if(count > len) break;
}
int left = temp - InOrder ;
int right = len - left - 1;
if(left > 0)
Rebuild(PostOrder, InOrder, left, root->left);
if(right > 0)
Rebuild(PostOrder + left, InOrder+left+1, right, root->right);
} int main()
{
int n,i,t;
while(cin>>n)
{
int *PostOrder = new int[n];
int *InOrder = new int[n];
for(i=0; i<n; i++)
cin>>PostOrder[i];
for(i=0; i<n; i++)
cin>>InOrder[i]; Node *root = new Node;
int post_start,in_start;
post_start = 0;
in_start = 0;
Rebuild(PostOrder, InOrder, n, root);
queue<Node *> q;
q.push(root);
int flag = 1; while(!q.empty()){
if(q.front()->left != NULL)
q.push(q.front()->left);
if(q.front()->right != NULL)
q.push(q.front()->right);
if(flag != n)
cout<<q.front()->value<<" ";
else
cout<<q.front()->value;
flag ++;
q.pop();
} cout<<endl;
}
return 0;
}

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