PAT 1115 Counting Nodes in a BST

 

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than or equal to the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

Insert a sequence of numbers into an initially empty binary search tree. Then you are supposed to count the total number of nodes in the lowest 2 levels of the resulting tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤) which is the size of the input sequence. Then given in the next line are the N integers in [ which are supposed to be inserted into an initially empty binary search tree.

Output Specification:

For each case, print in one line the numbers of nodes in the lowest 2 levels of the resulting tree in the format:

n1 + n2 = n

where n1 is the number of nodes in the lowest level, n2 is that of the level above, and n is the sum.

Sample Input:

9
25 30 42 16 20 20 35 -5 28

Sample Output:

2 + 4 = 6



建树，然后DFS找即可

struct node{
    int val;
    struct node *left,*right;
};
node* insert(node *root,int v){
    if(root==NULL){
        root=new node();
        root->val=v;
        root->left=NULL;
        root->right=NULL;
    }else if(v<=root->val){
        root->left=insert(root->left,v);
    }else{
        root->right=insert(root->right,v);
    }
    return root;
}
int maxd=0;
vector<int> v[1004];
void dfs(node *root,int dep){
    if(dep>maxd)maxd=dep;
    v[dep].push_back(root->val);
    if(root->left!=NULL)dfs(root->left,dep+1);
    if(root->right!=NULL)dfs(root->right,dep+1);
}
int main() {
    //freopen("D://dio.txt","r",stdin);
    int n,k,i;
    cin>>n;
    node *root=NULL;
    for(i=0;i<n;i++){
        scanf("%d",&k);
        root=insert(root,k);
    }
    dfs(root,1);
    printf("%d + %d = %d",v[maxd].size(),v[maxd-1].size(),v[maxd].size()+v[maxd-1].size());
    return 0;
}