1110 Complete Binary Tree (25分) 判断一棵二插树是否是完全二叉树

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 (≤) 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

解题思路：

这道题开始提交，有三个点通过不了，都显示段错误，想了半天，不应该段错误啊，原来是最开始输入，我用的是char输入，如果节点编号 >= 10那么char读不出来。应该用strng 输入，然后判断是否是" - "，是的话存为-1，否则正常存。

还有就是，判断完全二叉树，用层序遍历，如果前n个遍历序列没有出现-1，说明是完全二叉树，如果前n个序列出现了-1，则表示不是完全二叉树。代码如下

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>

using namespace std;
int tree[30];

int find(int x){  //寻找节点的根节点
	if(tree[x] == x){
		return x;
	}
	return find(tree[x]);
}

int main(){
	freopen("C:\\Users\\zzloyxt\\Desktop\\1.txt","r",stdin);
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++){  //存放每个下标节点的父亲节点。
		tree[i] = i;
	}
	int left[n];
	int right[n];
	char a[3],b[3];
	for(int i=0;i<n;i++){
		scanf("%s %s",a,b);
		if(strcmp(a,"-") ==0){
			left[i] = -1;
		}else{
			int a_int = 0;
			sscanf(a,"%d",&a_int);
			left[i] = a_int;
			tree[a_int] = i;
		}
		if(strcmp(b,"-") ==0){
			right[i] = -1;
		}else{
			int b_int = 0;
			sscanf(b,"%d",&b_int);  //将字符串转成整数。
			right[i] = b_int;
			tree[b_int] = i;
		}
	}
	int root = find(0);
	int shu[100] = {0};  //存放层序遍历序列
	int index = 0;
	queue<int> Q;
	Q.push(root);
	while(!Q.empty()){
		int x = Q.front();
		Q.pop();
		shu[index++] = x;
		if(x != -1){
			Q.push(left[x]);
			Q.push(right[x]);
		}
	}
	int flag = 0;
	for(int i=0;i<n;i++){
		if(shu[i] < 0){
			flag = 1;
			break;
		}
	}
	if(flag ==0){
		printf("YES %d\n",shu[n-1]);
	}else{
		printf("NO %d\n",root);
	}

	return 0;
}

原文地址：https://www.cnblogs.com/zzlback/p/12422430.html