1110 Complete Binary Tree

1110 Complete Binary Tree (25)（25 分）

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

 

题意：

给出树中每个结点的孩子结点（若无孩子，用-表示），要求判断其是否为完全二叉树。若是，输出YES及最后一个结点；若不是，输出NO和根结点。

 

思路：

利用层序遍历，把树的所有结点（包括空结点）都push进队列。设立全局变量cnt，初始化cnt=0，用来记录已经访问到的非空结点的个数，在遇到第一个空结点时，若cnt==n，则是完全二叉树；若cnt<n，则不是完全二叉树。

 

如下图，该图为完全二叉树，则在队列中元素的存储是这样的：

null(4r) null(4l) null(3r) null(3l) null(2r) 4 3 2 1

而对于如下这棵树，不是完全二叉树，则在队列中元素的存储是这样的：

null(4r) null(4l)  null(3l) null(2r) null(2l) 3 2 1

 

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
using namespace std;

struct Node{
    int left,right;
}Tree[];
int n;

bool isCBT(int root,int &lastval)
{
    ;
    queue<int> q;
    q.push(root);
    while(!q.empty()){
        int top=q.front();
        q.pop();
        ){
            cnt++;
            lastval=top;
            q.push(Tree[top].left);
            q.push(Tree[top].right);
        }else {
            if(cnt==n) return true;
            else return false;
        }
    }
}

int main()
{
    scanf("%d",&n);
    ],u[];
    int left,right;
    ];
    memset(isRoot,true,sizeof(isRoot));
    ;i<n;i++){
        scanf("%s %s",v,u);
        ]==;
        else{
            left=atoi(v);
            isRoot[left]=false;
        }
        ]==;
        else{
            right=atoi(u);
            isRoot[right]=false;
        }
        Tree[i].left=left;
        Tree[i].right=right;
    }
    ;
    while(isRoot[root]==false) root++;
    ;
    bool flag=isCBT(root,lastVal);
    if(flag) printf("YES %d\n",lastVal);
    else printf("NO %d\n",root);
    ;
}