【PAT】A1110 Complete Binary Tree (25)

最新推荐文章于 2024-05-06 22:07:52 发布

ztmajor

最新推荐文章于 2024-05-06 22:07:52 发布

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本文介绍了一种用于判断给定树是否为完全二叉树的算法。通过深度优先搜索，算法确定树的最大叶子节点位置，进而判断树的类型。文章详细解释了输入输出规格，并提供了代码实现。

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1110 Complete Binary Tree (25 point(s))

1110 Complete Binary Tree (25 point(s))

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

Code

#include <iostream>
//#include <fstream>
using namespace std;
struct node{
    int l, r;
}nodes[21];							// N <= 20
int maxindex = -1, res;				// maxindex:最大的叶子位置；res:存放根节点序号
void dfs(int root, int index) {		// 深度优先搜索，找到最大的index，找到树的最后一个节点
    if(index > maxindex) {
        maxindex = index;
        res = root;
    }
    if(nodes[root].l != -1) dfs(nodes[root].l, index * 2);
    if(nodes[root].r != -1) dfs(nodes[root].r, index * 2 + 1);
}
int main() {
    //ifstream cin("/home/my/Desktop/in.txt");
    int n, root = 0, haveroot[100] = {0};	// haveroot[]，是否有父节点
    cin >> n;
    string l, r;
    for (int i = 0; i < n; i++) {			// 接收输入，存放节点的左右子树
        cin >> l >> r;
        if (l == "-") {
            nodes[i].l = -1;
        } else {
            nodes[i].l = stoi(l);
            haveroot[stoi(l)] = 1;
        }
        if (r == "-") {
            nodes[i].r = -1;
        } else {
            nodes[i].r = stoi(r);
            haveroot[stoi(r)] = 1;
        }
    }
    while (haveroot[root] != 0) root++;	// 找到树的根节点
    dfs(root, 1);
    if (maxindex == n)      printf("YES %d", res);
    else                    printf("NO %d", root);
    return 0;
}