标签：node Binary PAT Complete int s1 tree st step

https://pintia.cn/problem-sets/994805342720868352/problems/994805359372255232

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 1

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
int vis[110];
int ans = -1, temp;

struct Node{
    int l;
    int r;
}node[110];

int StringtoInt(string s) {
    int len = s.length();
    int sum = 0;
    for(int i = 0; i < len; i ++) {
        sum = sum * 10 + (s[i] - '0');
    }
    return sum;
}

void dfs(int st, int step) {
    if(step > ans) {
        ans = step;
        temp = st;
    }

    if(node[st].l != -1) dfs(node[st].l, step * 2);
    if(node[st].r != -1) dfs(node[st].r, step * 2 + 1);
}

int main() {
    scanf("%d", &N);
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < N; i ++) {
        string s1, s2;
        cin >> s1 >> s2;
        if(s1 == "-") {
            node[i].l = -1;
        } else if(s1 != "-"){
            node[i].l = StringtoInt(s1);
            //cout << node[i].l << endl;
            vis[node[i].l] = 1;
        }

        if(s2 == "-") {
            node[i].r = -1;
        } else if(s2 != "-") {
            node[i].r = StringtoInt(s2);
            vis[node[i].r] = 1;
            //cout << node[i].r << endl;
        }
    }

    int root = 0;
    while(vis[root]) root ++;
    dfs(root, 1);

    if(ans == N)
        printf("YES %d\n", temp);
    else printf("NO %d\n", root);
    return 0;
}

　　判断是不是完全二叉树要判断这个树是不是满的 看是不是最大的节点数等于一共的节点数目 dfs 求最大的节点下标

FH

标签：node,Binary,PAT,Complete,int,s1,tree,st,step
来源： https://www.cnblogs.com/zlrrrr/p/10391420.html

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