1110 Complete Binary Tree(25 分)
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
题目大意:给出一棵树二叉树,判断是否是完全二叉树,如果是那么输出最后一个节点;如果不是输出根节点。
//第一次见完全二叉树的题目,想起了完全二叉树的性质,存储树的话,就用结构体数组,下标表示当前节点号;首先求出树的高度根据logn,看是否余数为0,判断是否+1;那么前n-1层的节点要是满的,并且再通过只有一个左子节点或者右子节点的树只有一个,那么来判断是否是完全二叉树;并且结构体里有一个属性是father默认为-1。感觉好复杂,就没有用代码实现。
代码来自:https://www.liuchuo.net/archives/2158
#include <iostream> #include <queue> #include <vector> #include <string> using namespace std; struct TREE { int left, right; }; int main() { int n, root = 0; scanf("%d", &n); vector<TREE> tree(n); vector<int> book(n); for(int i = 0; i < n; i++) { string l, r; cin >> l >> r;//使用字符串读取,也必须使用字符串, if(l == "-") { tree[i].left = -1;//如果左右为空的话,则标记为-1. } else { tree[i].left = stoi(l);//不用使用-'0'将其转换,直接使用stoi函数即可 book[tree[i].left] = 1; } if(r == "-"){ tree[i].right = -1; } else { tree[i].right = stoi(r); book[tree[i].right] = 1; } } for(int i = 0; i < n; i++) { if(book[i] == 0) { root = i; break;//没有出现的便是根! } } queue<int> q; q.push(root); int cnt = 0, lastnode = 0; while(!q.empty()) { int node = q.front(); q.pop(); if(node != -1) { lastnode = node; cnt++;//记录层次遍历在-1出现之前的节点数 }else { if(cnt != n) printf("NO %d", root); else printf("YES %d", lastnode); return 0; } q.push(tree[node].left);//如果左右子节点为空,那么就将-1push进去了 q.push(tree[node].right); } return 0; }
//学习了!
1.根据输入建树,每个节点因为本身就是ID,左右如果是空节点,那么就赋值为-1.
2.根节点是怎么找到的呢?在建树输入的过程中,如果一个点没有出现,那么就是根节点,因为都在一棵树中!都是表示的是子节点,如果没出现,就表示它不是子节点,而是根节点!
3.如何去判断是否是CBT呢?使用层次遍历!并且记录当前层次遍历的个数,根据CBT的性质,如果当前出现空节点,但是遍历过的点数!=总结点数,那么就不是二叉树,可以画一个图试试!使用队列!
//学习了!