PAT-A1110. 二叉树-完全二叉树的判断

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;
}