[二叉树建树&完全二叉树判断] 1110. Complete Binary Tree (25)

1110. Complete Binary Tree (25)

Given a tree, you are supposed to tell if it is a complete binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=20) 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, 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 case, print in one line "YES" and the index of the last node if the tree is a complete binary tree, or "NO" and the index of the root if not. There must be exactly one space separating the word and the number.

Sample Input 1:

9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -

Sample Output 1:

YES 8

Sample Input 2:

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

Sample Output 2:

NO 1

分析：这道题目的建树与以前的通过中序和前序或者中序和后序建树不同，这里给出了每个孩子的左右孩子，根据这些信息来建树。做法是申明一个节点数组，将对应的节点信息填入数组即可，而左右孩子指针就是数组的下标。另外，关于完全二叉树的判断，网上也有很多资料。

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

struct Node
{
    int data;
    int lchild,rchild;
}node[10000];

int inDreeg[10000]={0};

bool judge(int root,int & last_node)
{
    if(root==-1) return true;
    queue<int> q;
    q.push(root);
    int flag=0;
    int now;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        //cout<<node[now].data<<endl;
        if(flag==0)
        {
            if(node[now].lchild!=-1&&node[now].rchild!=-1) flag=0;
            else if(node[now].lchild==-1&&node[now].rchild!=-1) return false;
            else flag=1;
        }
        else
        {
            if(node[now].lchild!=-1||node[now].rchild!=-1) return false;
        }
        if(node[now].lchild!=-1) q.push(node[now].lchild);
        if(node[now].rchild!=-1) q.push(node[now].rchild);
    }
    last_node=node[now].data;
    return true;
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        string l,r;
        cin>>l>>r;
        node[i].data=i;
        if(l=="-")
        {
            node[i].lchild=-1;
        }
        else
        {
            int id=atoi(&l[0]);
            node[i].lchild=id;
            inDreeg[id]+=1;
        }
        
        if(r=="-")
        {
            node[i].rchild=-1;
        }
        else
        {
            int id=atoi(&r[0]);
            node[i].rchild=id;
            inDreeg[id]+=1;
        }
    }
    bool ans;
    int root=-1;
    for(int i=0;i<n;i++)
    {
        if(inDreeg[i]==0)
        {
            root=i;
            break;
        }
    }
    int last;
    ans=judge(root,last);
    if(ans==true)
    {
        printf("YES %d\n",last);
    }
    else
    {
        printf("NO %d\n",root);
    }
}