1020 Tree Traversals (25 分)

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

代码： 

#include<iostream>
#include<queue>
using namespace std;
struct node
{
    int data;
    node* lchild,*rchild;
};
int post[35],in[35];
void levelorder(node *root)//层序遍历
{
    if(root==NULL)
        return;
    queue<node *> Q;
    Q.push(root);
    while(!Q.empty())
    {
        node *temp=Q.front();
        if(temp==root)
            printf("%d",root->data);
        else
            printf(" %d",temp->data);
        Q.pop();
        if(temp->lchild)
            Q.push(temp->lchild);
        if(temp->rchild)
            Q.push(temp->rchild);
    }
}
node* creatree(int postL,int postR,int inL,int inR)//根据后序遍历和中序遍历结果，创建二叉树
{
    if(postL>postR)
        return NULL;
    int temp=post[postR];
    node* root=new node;
    root->data=temp;
    int k;
    for(k=inL;k<=inR;k++)
        if(in[k]==temp)
            break;
    int newk=k-inL;
    root->lchild=creatree(postL,postL+newk-1,inL,inL+newk-1);
    root->rchild=creatree(postL+newk,postR-1,k+1,inR);
    return root;
}
int main()
{
    int N;
    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=creatree(0,N-1,0,N-1);
    levelorder(root);
    return 0;
}

标签：node,25,temp,int,Traversals,Tree,postL,line,root
来源： https://blog.csdn.net/fairylyt/article/details/123136884