PAT_A1110#Complete Binary Tree

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/blue-lin/p/11032261.html

本文介绍了一种算法，用于判断给定的树是否为完全二叉树。通过输入节点数量及各节点的左右子节点信息，算法使用队列进行层次遍历，检查树的结构是否符合完全二叉树的定义。

Source:

PAT A1110 Complete Binary Tree (25 分)

Description:

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

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

Keys:

完全二叉树（Complete Binary Tree）

Attention:

atoi(s.c_str()); 字符串转化为int，atof 对应 double，atoll 对应 long long；
含非数字则截取前面的数字部分，首字符非数字则返回0

Code:

 1 #include<cstdio>
 2 #include<string>
 3 #include<queue>
 4 #include<iostream>
 5 using namespace std;
 6 const int M=25;
 7 int h[M]={0};
 8 struct node
 9 {
10     int left,right;
11 }tree[M];
12 
13 bool isComplete(int root, int &last)
14 {
15     queue<int> q;
16     q.push(root);
17     while(!q.empty())
18     {
19         root = q.front();
20         q.pop();
21         if(root != -1)
22         {
23             last = root;
24             q.push(tree[root].left);
25             q.push(tree[root].right);
26         }
27         else
28         {
29             while(!q.empty())
30             {
31                 if(q.front() != -1)
32                     return false;
33                 q.pop();
34             }
35         }
36     }
37     return true;
38 }
39 
40 int main()
41 {
42 #ifdef    ONLINE_JUDGE
43 #else
44     freopen("Test.txt", "r", stdin);
45 #endif
46 
47     int n,root,last;
48     scanf("%d", &n);
49     for(int i=0; i<n; i++)
50     {
51         string l,r;
52         cin >> l >> r;
53         if(l == "-")
54             tree[i].left=-1;
55         else{
56             tree[i].left = atoi(l.c_str());
57             h[tree[i].left]=1;
58         }
59         if(r == "-")
60             tree[i].right=-1;
61         else{
62             tree[i].right = atoi(r.c_str());
63             h[tree[i].right]=1;
64         }
65     }
66     for(int i=0; i<n; i++){
67         if(h[i]==0){
68             root=i;
69             break;
70         }
71     }
72     if(isComplete(root, last))
73         printf("YES %d", last);
74     else
75         printf("NO %d", root);
76 
77     return 0;
78 }