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 11 #include <iostream> 2 #include <vector> 3 #include <queue> 4 #include <string> 5 using namespace std; 6 struct Node 7 { 8 int l, r; 9 }node; 10 int n; 11 int main() 12 { 13 cin >> n; 14 vector<Node>tree; 15 vector<bool>isRoot(n, true); 16 for (int i = 0; i < n; ++i) 17 { 18 string s1, s2; 19 cin >> s1 >> s2; 20 if (s1 == "-") 21 node.l = -1; 22 else 23 { 24 node.l = atoi(s1.c_str()); 25 isRoot[node.l] = false; 26 } 27 if (s2 == "-") 28 node.r = -1; 29 else 30 { 31 node.r = atoi(s2.c_str()); 32 isRoot[node.r] = false; 33 } 34 tree.push_back(node); 35 } 36 int root = -1;//根 37 for (int i = 0; i < n && root==-1; ++i) 38 if (isRoot[i]) 39 root = i; 40 queue<int>q, temp; 41 q.push(root); 42 while (!q.empty())//进行层序遍历 43 { 44 int p = q.front(); 45 q.pop(); 46 temp.push(p);//保存弹出的数据 47 if (tree[p].l != -1) 48 q.push(tree[p].l); 49 else 50 break;//出现空子节点，则打破了完全二叉树的规则 51 if (tree[p].r != -1) 52 q.push(tree[p].r); 53 else 54 break;//出现空子节点，则打破了完全二叉树的规则 55 } 56 if (temp.size() + q.size() == n)//满足完全二叉树的要求 57 cout << "YES " << q.back() << endl;//最后压入的就是最后一个节点 58 else 59 cout << "NO " << root << endl; 60 return 0; 61 }