1110. Complete Binary Tree （判断完全二叉树）

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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:

97 8- -- -- -0 12 34 5- -- -

Sample Output 1:

YES 8

Sample Input 2:

8- -4 50 6- -2 3- 7- -- -

Sample Output 2:

NO 1

给你一颗二叉树，让你判断是否为完全二叉树...

方法一：给每个节点标记一下，例如，root=1，lson=2，rson=3....

最后判断最后一个节点的值是否等于N

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int N = 22;char s1[5],s2[5];int pre[N],have[N];queue<int>qu;struct node{int l,r,num;}a[N];int main(){int i,j,n,l,root,flag=1,f,r;scanf("%d",&n);memset(pre,-1,sizeof(pre));for(i=0;i<n;i++) {scanf("%s%s",s1,s2);l=r=-1;if(s1[0]!='-') {sscanf(s1,"%d",&l);pre[l]=i;}if(s2[0]!='-') {sscanf(s2,"%d",&r);pre[r]=i;}a[i].l=l,a[i].r=r;}root=0;while(pre[root]!=-1) root=pre[root];a[root].num=1;qu.push(root);     while(!qu.empty()) {    f=qu.front();    qu.pop();    l=a[f].l,r=a[f].r;    a[l].num=a[f].num<<1;    a[r].num=a[f].num<<1|1;    if(l!=-1) qu.push(l);    if(r!=-1) qu.push(r);}if(a[f].num==n) printf("YES %d\n",f);else printf("NO %d\n",root);return 0;}

方法二是，遍历的时候把左右子树都扔进去，直到遍历到NULL时退出来，之后再判断一个队列里面有没有还有没遍历到的点

任意的一个二叉树，都可以补成一个满二叉树。这样中间就会有很多空洞。在广度优先遍历的时候，如果是满二叉树，或者完全二叉树，这些空洞是在广度优先的遍历的末尾，所以，但我们遍历到空洞的时候，整个二叉树就已经遍历完成了。而如果，是非完全二叉树，

我们遍历到空洞的时候，就会发现，空洞后面还有没有遍历到的值。这样，只要根据是否遍历到空洞，整个树的遍历是否结束来判断是否是完全的二叉树。

bool is_complete(tree *root)
{
   queue q;
   tree *ptr;
   // 进行广度优先遍历（层次遍历），并把NULL节点也放入队列
   q.push(root);
   while ((ptr = q.pop()) != NULL)
   {
       q.push(ptr->left);
       q.push(ptr->right);
   }

   // 判断是否还有未被访问到的节点
   while (!q.is_empty())
   {
       ptr = q.pop();

       // 有未访问到的的非NULL节点，则树存在空洞，为非完全二叉树
       if (NULL != ptr)
       {
           return false;
       }
   }

   return true;
}

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