PAT-A1110. 二叉树-完全二叉树的判断

最新推荐文章于 2022-11-25 10:16:06 发布

lilysara

最新推荐文章于 2022-11-25 10:16:06 发布

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分类专栏： PAT 文章标签：二叉树

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本文介绍了一种通过层序遍历来判断一棵树是否为完全二叉树的方法，并提供了一个具体的C++实现示例。完全二叉树是指除了最后一层外，每一层都是满的，并且最后一层的节点都集中在该层左边的二叉树。

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题目链接：https://www.patest.cn/contests/pat-a-practise/1110
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

完全二叉树判断思路：层序遍历，一个完全二叉树中，当遍历到某个结点没有子结点时，那么它后面的所有结点都应该没有子结点。

#include <iostream>
#include <queue>
#include <cstring>
#include <string>
using namespace std;

struct BTree {
    int lchild,rchild;
};
BTree T[25];
bool notRoot[25];
int thisone = -1; // 层序遍历时当前访问的节点序数

bool LevelOrder(int root)
{
    queue<int> qt;
    qt.push(root);
    bool nochild = false;
    while (!qt.empty()) {
        thisone = qt.front();
        qt.pop();
        if(T[thisone].lchild!=-1){
            if(nochild)
                return false;
            qt.push(T[thisone].lchild);
        }
        else
            nochild = true;
        if(T[thisone].rchild!=-1){
            if(nochild)
                return false;
            qt.push(T[thisone].rchild);
        }
        else
            nochild = true;
    }
    return true;
}

int GetRootIndex()
{
    int i = 0;
    while (notRoot[i]) {
        ++i;
    }
    return i;
}

int main()
{
    memset(notRoot, 0, sizeof(notRoot));
    int num;
    cin >> num;
    for(int i = 0; i < num; ++i)
    {
        char str[3];
        int temp;
        scanf("%s", str);
        if(str[0] == '-') T[i].lchild = -1;
        else {
            sscanf(str, "%d", &temp);
            T[i].lchild = temp;
            notRoot[temp] = true;
        }

        scanf("%s", str);
        if(str[0] == '-') T[i].rchild = -1;
        else {
            sscanf(str, "%d", &temp);
            T[i].rchild = temp;
            notRoot[temp] = true;
        }

    }

    int rootIndex = GetRootIndex();

    if(LevelOrder(rootIndex))
        cout << "YES" << " " << thisone << endl;
    else
        cout << "NO" << " " << rootIndex << endl;

    return 0;
}