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
题意：给出 n 个结点,（0-n-1）n 行左右孩子结点，根结点为行中没有出现的结点；
判断是否为完全二叉树，是则输出 (YES) 与末尾结点，否则输出(NO)与根结点。
AC【判断完全二叉树】：1、[ 找出根结点 ] 2、[ DFS遍历 ] 3、[ 分类输出 ]
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 struct node{ 5 string left, right; 6 } Node[21]; 7 8 int root= 0, maxn = -1, last; 9 10 void DFS(int root, int index) 11 { 12 if(index > maxn) 13 { 14 maxn = index; 15 last = root; 16 } 17 if(Node[root].left != "-") DFS(stoi(Node[root].left), index * 2); 18 if(Node[root].right != "-") DFS(stoi(Node[root].right), index * 2 + 1); 19 } 20 21 int main() 22 { 23 int n; cin >> n; 24 vector<int> find_root(n, 0); 25 for(int i = 0; i < n; i++) 26 { 27 cin >> Node[i].left >> Node[i].right; 28 if(Node[i].left != "-") find_root[stoi(Node[i].left)] = 1; 29 if(Node[i].right != "-") find_root[stoi(Node[i].right)] = 1; 30 } 31 32 while(root < n && find_root[root]) root++; 33 DFS(root, 1); 34 35 if(n == maxn) cout << "YES " << last; 36 else cout << "NO " << root; 37 return 0; 38 }
Tips：vector<int> v(n, 0); // 初始化