[递归&&bfs]PAT1020 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:

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

Sample Output:

4 1 6 3 5 7 2

题意:根据题中所给出的后序遍历和中序遍历序列求树的层次遍历。

思路：先建立二叉树，再使用bfs就可以求解。

#include<iostream>
#include<queue>

using namespace std;

int postorder[40],inorder[40];

class Node
{
public:
    int data;
    Node *ld,*rd;
};

queue<Node*> st;

Node* CreateTree(int n,int *arry1,int *arry2)
{
    if(n<=0) return NULL;
    int k=0;
    while(arry1[n-1]!=arry2[k])
        k++;
    Node *root=new Node;
    root->data=arry1[n-1];
    root->ld=CreateTree(k,arry1,arry2);
    root->rd=CreateTree(n-k-1,arry1+k,arry2+k+1);
    return root;
}


int main()
{
    int n,i,j,k;
    cin>>n;
    for(i=0;i<n;i++)
        cin>>postorder[i];
    for(i=0;i<n;i++)
        cin>>inorder[i];
    Node *root=new Node;
    root->ld=NULL;
    root->rd=NULL;
    root=CreateTree(n,postorder,inorder);
    st.push(root);
    //cout<<(st.back())->data<<endl;
    int tag=0;
    while(!st.empty())
    {
        if((st.front())->ld!=NULL) st.push((st.front())->ld);
        if((st.front())->rd!=NULL) st.push((st.front())->rd);
        if(tag==0) cout<<(st.front())->data,tag=1;
        else cout<<" "<<(st.front())->data;
        st.pop();
    }
    return 0;
}