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 (≤), 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
#include<stdio.h>
#include<queue>
using namespace std;
int num;
int postorder[50];
int inorder[50];
struct Tree
{
Tree * leftchild;
Tree * rightchild;
int date;
};
queue<int> out;
Tree * findtree(int postL,int postR,int inL,int inR)
{
if(postL>postR) //postl=postr说明还有一个单个的
{
return NULL;
}
Tree *root=new Tree;
root->date=postorder[postR];
int k;
for(k=inL;k<inR;k++)
{
if(inorder[k]==postorder[postR])
{
break;
}
}
int numLeft=k-inL;
root->leftchild=findtree(postL,postL+numLeft-1,inL,k-1);
root->rightchild=findtree(postL+numLeft,postR-1,k+1,inR); //postr位为根
return root;
}
int n=0;
void bfs(Tree* root)
{
queue<Tree*> q;
q.push(root);
while(!q.empty())
{
Tree * now=q.front();
q.pop();
if(n!=num-1)
{
printf("%d ",now->date);
n++;
}
else
{
printf("%d",now->date);
}
if(now->leftchild!=NULL)
{
q.push(now->leftchild);
}
if(now->rightchild!=NULL)
{
q.push(now->rightchild);
}
}
}
int main()
{
scanf("%d",&num);
for(int i=0;i<num;i++)
{
scanf("%d",&postorder[i]);
}
for(int i=0;i<num;i++)
{
scanf("%d",&inorder[i]);
}
Tree * root=findtree(0,num-1,0,num-1);
bfs(root);
}根据先序中序建树，bfs用queue实现即可
