题目描述：
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 8Sample Input 2:
8 - - 4 5 0 6 - - 2 3 - 7 - - - -Sample Output 2:
NO 1
题目思路：
大概意思：题目给出了n个，并分别给出了n个结点的左右孩子。题目要求判断该树是否是完全二叉树。如果是，输出YES和最后一个结点的编号，否则输出NO并且输出根节点的编号。
用结构体v[i].l和v[i].r分别表示i结点的左孩子和右孩子，并且对孩子结点标记。dfs搜索所有的结点一遍，没有被标记的结点便是根节点。
找到根节点后，我们可以层次遍历整棵树，如果搜索到孩子结点为-1的情况，则比较当前位置的编号和n的关系。如果相等，则是完全二叉树，否则不是。
题目代码：
#include <cstdio>
#include <vector>
#include <iostream>
#include <queue>
using namespace std;
int n, temp, cnt, lastnode, root;
string s1, s2;
// 树
struct Tree{int l,r;
};int main(){cin>>n;vector<Tree>v(n); // 用来存放树 vector<int>m(n);for(int i = 0; i < n; i++){cin>>s1>>s2;if(s1[0] != '-'){temp = s1[0] - '0';if(s1.length() == 2){temp = s1[1] - '0' + temp * 10;}v[i].l = temp; // lc}else{v[i].l = -1;}if(s2[0] != '-'){temp = s2[0] - '0';if(s2.length() == 2){temp = s2[1] - '0' + temp * 10;}v[i].r = temp; // rc }else{v[i].r = -1;}// m[i] i不能是负数 否则会异常退出 if(v[i].l != -1)m[v[i].l] = 1;if(v[i].r != -1)m[v[i].r] = 1;}// 寻找根节点 for(int i = 0; i < n; i++){if(m[i] == 0){root = i;break;}}// 层次遍历 queue<int>q;cnt = 0; lastnode = 0; q.push(root);while(!q.empty()){int node = q.front();if(node != -1){cnt++;lastnode = node;}else{if(cnt == n)printf("YES %d\n",lastnode);elseprintf("NO %d\n",root);break;}q.push(v[node].l);q.push(v[node].r);q.pop();}return 0;
}