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
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<string.h> 5 #include<queue> 6 using namespace std; 7 typedef struct{ 8 int lchild, rchild; 9 }node; 10 node tree[21]; 11 int str2num(char ss[]){ 12 int len = strlen(ss); 13 if(ss[0] == '-') 14 return -1; 15 int P = 1, ans = 0; 16 for(int i = len - 1; i >= 0; i--){ 17 ans += (ss[i] - '0') * P; 18 P *= 10; 19 } 20 return ans; 21 } 22 int N, lastNode, root, notRoot[21] = {0}, cnt = 0, tag = 1; 23 void levelOrder(int root){ 24 queue<int> Q; 25 Q.push(root); 26 while(Q.empty() == false){ 27 int temp = Q.front(); 28 lastNode = temp; 29 Q.pop(); 30 cnt++; 31 if(cnt < N / 2 && (tree[temp].lchild == -1 || tree[temp].rchild == -1)){ 32 tag = 0; 33 return; 34 }else if(cnt == N / 2 && (tree[temp].lchild == -1 && tree[temp].rchild == -1 || tree[temp].lchild == -1 && tree[temp].rchild != -1)){ 35 tag = 0; 36 return; 37 }else if(cnt > N / 2 && (tree[temp].lchild != -1 || tree[temp].rchild != -1)){ 38 tag = 0; 39 return; 40 } 41 if(tree[temp].lchild != -1) 42 Q.push(tree[temp].lchild); 43 if(tree[temp].rchild != -1) 44 Q.push(tree[temp].rchild); 45 } 46 } 47 int main(){ 48 scanf("%d", &N); 49 char str[10]; 50 for(int i = 0; i < N; i++){ 51 scanf("%s", str); 52 tree[i].lchild = str2num(str); 53 scanf("%s", str); 54 tree[i].rchild = str2num(str); 55 if(tree[i].lchild != -1) 56 notRoot[tree[i].lchild] = 1; 57 if(tree[i].rchild != -1) 58 notRoot[tree[i].rchild] = 1; 59 } 60 for(int i = 0; i < N; i++){ 61 if(notRoot[i] == 0){ 62 root = i; 63 break; 64 } 65 } 66 levelOrder(root); 67 if(tag == 0) 68 printf("NO %d", root); 69 else printf("YES %d", lastNode); 70 cin >> N; 71 return 0; 72 }
总结：
1、题目要求判断二叉树是否是完全二叉树。我的办法是二叉树的层序遍历，访问一个节点就cnt++。访问前 N/2 - 1个节点时要求必须都有左右孩子。第N/2个节点要求必须是左右都非空或左非空右为空。N/2之后的节点要求左右子树都必须空。
2、网上看到还有更简单的方法，就是在层序遍历的时候把为-1的空节点也都加入队列。当访问时，遇到-1节点则查看cnt，如果cnt<N 则说明不是完全二叉树。另外注意，最后一个非空节点不一定是N-1。因为虽然是完全二叉树，但它的层次遍历节点序号不是按照0、1、2、3的顺序。