• 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 (≤), 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

    基本思路:
      1、因为中序遍历和后序遍历可以唯一确定一棵二叉树,所以首先构造处这颗树,然后进行层序遍历即可
      2、建立二叉树的时候要判断左右子树长度为0,并且要注意参数的值传递。具体如何做可参考代码。
      
    #include<iostream>
    #include<sstream>
    #include<algorithm>
    #include<map>
    #include<cmath>
    #include<cstdio>
    #include<string.h>
    #include<queue>
    #include<vector>
    using namespace std;
    struct Node
    {
        int data;
        Node *lchild,*rchild;
    };
    
    void CreateTree(int inOrder[],int iL,int iR,
                     int postOrder[],int pI,int pR,Node *&root)
    {
        int temp=0;
        while(inOrder[temp]!=postOrder[pR]) temp++;
        int leftLength=temp-iL;
        int rightLength=iR-temp;
        if(root==nullptr)
        {
            root =new Node();
            root->data=postOrder[pR];
        }
        if(leftLength>0)
            CreateTree(inOrder,iL,temp-1,postOrder,pI,pR-rightLength-1,root->lchild);
        if(rightLength>0)
            CreateTree(inOrder,temp+1,iR,postOrder,pI+leftLength,pR-1,root->rchild);
    
    }
    int a[31];
    void level(Node *root)
    {
        queue<Node*> q;
        q.push(root);
        int i=0;
        while(!q.empty())
        {
            Node *temp=q.front();
            q.pop();
            a[i++]=temp->data;
            if(temp->lchild)
                q.push(temp->lchild);
            if(temp->rchild)
                q.push(temp->rchild);
        }
    }
    
    int main()
    {
        int n;
        cin>>n;
        int postOrder[n],inOrder[n];
        for(int i=0; i<n; i++)
            cin>>postOrder[i];
        for(int i=0; i<n; i++)
            cin>>inOrder[i];
        Node *root=nullptr;
        CreateTree(inOrder,0,n-1,postOrder,0,n-1,root);
        level(root);
    
        if(n>0)
            cout<<a[0];
        for(int i=1;i<n;i++)
            cout<<" "<<a[i];
    
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/zhanghaijie/p/10228464.html
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