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  • PAT 1102 Invert a Binary Tree

    The following is from Max Howell @twitter:

    Google: 90% of our engineers use the software you wrote (Homebrew), but you can't invert a binary tree on a whiteboard so fuck off.

    Now it's your turn to prove that YOU CAN invert a binary tree!

    Input Specification:

    Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node from 0 to N-1, and gives the indices of the left and right children of the node. If the child does not exist, a "-" will be put at the position. Any pair of children are separated by a space.

    Output Specification:

    For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

    Sample Input:

    8
    1 -
    - -
    0 -
    2 7
    - -
    - -
    5 -
    4 6

    Sample Output:

    3 7 2 6 4 0 5 1
    6 5 7 4 3 2 0 1

    分析
    用vector Left和right来记录每个节点的左孩子节点和右孩子节点,利用根节点不是任何节点的子节点来找到根节点root,在从根节点开始将节点依次插入树中,最后就是基本的层序遍历和中序遍历的实现了。

    #include<iostream>
    #include<vector>
    #include<queue>
    #include<string.h>
    using namespace std;
    vector<int> Left;
    vector<int> Right;
    int flag=1;
    class node{
      public:
        int value;
        node* left=NULL;
        node* right=NULL;
        node(int val):value(val), left(NULL), right(NULL){
        }
    };
    void CreateTree(node* root){
      if(Left[root->value]!=-1){
        root->left=new node(Left[root->value]);
        CreateTree(root->left);
      }
      if(Right[root->value]!=-1){
        root->right=new node(Right[root->value]);
        CreateTree(root->right);
      }
    }
    void inorder(node* root){
      if(root->left)
        inorder(root->left);
      flag++==1?cout<<root->value:cout<<" "<<root->value;
      if(root->right)
        inorder(root->right);
    }
    int main(){
      int N;
      cin>>N;
      vector<int> record(N,0);
      Left.resize(N);
      Right.resize(N);
      for(int i=0; i<N; i++){
        char l, r;
        cin>>l>>r;
        if(isdigit(l)){
          Right[i]=l-'0';
          record[Right[i]]=1;
        }
        else
          Right[i]=-1;
        if(isdigit(r)){
          Left[i]=r-'0';
          record[Left[i]]=1;
        }
        else
          Left[i]=-1;
      }
      int i;
      for(i=0; i<N; i++)
        if(record[i]==0)
          break;
      node* root=new node(i);
      CreateTree(root);
      queue<node*> q;
      q.push(root);
      while(q.size()!=0){
        node* t=q.front();
        q.pop();
        flag++==1?cout<<t->value:cout<<" "<<t->value;
        if(t->left)
          q.push(t->left);
        if(t->right)
          q.push(t->right);
      }
      cout<<endl;
      flag=1;
      inorder(root);
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/9501849.html
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