1110 Complete Binary Tree

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最新推荐文章于 2021-03-08 20:20:56 发布

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分类专栏： PAT甲级题解文章标签： pat 完全二叉树

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本文介绍了一种判断一棵树是否为完全二叉树的方法，通过构建树并给每个节点编号，比较最大编号与节点总数来确定。文章提供了C++实现代码，包括树的结构定义、广度优先搜索算法及主函数。

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1110 Complete Binary Tree
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
分析：判断一棵树是否为完全二叉树。
先构建一棵树，然后给树的每个结点进行编号，根结点为1，左孩子结点的编号为2，有孩子结点的编号为3，以此类推，最大编号maxn与节点个数n进行比较，如果两者相等，则此树是完全二叉树，反之则不是。
参考代码（来自别处）

#include<iostream>
#include<string>
#include<vector>
using namespace std;
struct node {
    int left, right;
};
vector<node>tree;
int maxn = 0, ans = 0;
void bfs(int vec,int index) {
    if (index > maxn) {
        maxn= index;
        ans = vec;
    }
    if (tree[vec].left != -1)bfs(tree[vec].left, 2 * index);
    if (tree[vec].right != -1)bfs(tree[vec].right, 2 * index+1);
}
int main()
{
    int n, root=0;
    scanf_s("%d", &n);
    vector<int>v;
    v.resize(n, -1);
    string l, r;
    tree.resize(n);
    getchar();
    for (int i = 0; i < n; i++) {
        cin >> l >> r;
        if (l == "-")tree[i].left = -1;
        else {
            tree[i].left = stoi(l);
            v[stoi(l)] = 1;
        }
        if (r == "-")tree[i].right = -1;
        else {
            tree[i].right = stoi(r);
            v[stoi(r)] = 1;
        }
    }
    while (v[root]!=-1) { root++; }
    bfs(root, 1);
    if (maxn== n)cout << "YES" << " " << ans;
    else cout << "NO" << " " << root;

    return 0;
}