A1115. 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 (<=1000) which is the size of the input sequence. Then given in the next line are the N integers in [-1000 1000] 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

 #include<cstdio>

 #include<iostream>

 #include<algorithm>

 #include<string>

 #include<queue>

 using namespace std;

 typedef struct NODE{

     struct NODE *lchild, *rchild;

     int data;

     int level;

 }node;

 void insert(node* &root, int x){

     if(root == NULL){

         node* temp = new node;

         temp->lchild = NULL;

         temp->rchild = NULL;

         temp->data = x;

         root = temp;

         return;

     }

     if(x <= root->data)

         insert(root->lchild, x);

     else insert(root->rchild, x);

 }

 int depth = , n1, n2;

 void levelOrder(node* root){

     queue<node*> Q;

     root->level = ;

     Q.push(root);

     while(Q.empty() == false){

         node* temp = Q.front();

         depth = temp->level;

         Q.pop();

         if(temp->lchild != NULL){

             temp->lchild->level = temp->level + ;

             Q.push(temp->lchild);

         }

         if(temp->rchild != NULL){

             temp->rchild->level = temp->level + ;

             Q.push(temp->rchild);

         }

     }

 }

 void DFS(node * root){

     if(root == NULL)

         return;

     if(root->level == depth)

         n1++;

     else if(root->level == depth -)

         n2++;

     DFS(root->lchild);

     DFS(root->rchild);

 }

 int main(){

     int N, temp;

     node* root = NULL;

     scanf("%d", &N);

     for(int i = ; i < N; i++){

         scanf("%d", &temp);

         insert(root, temp);

     }

     levelOrder(root);

     DFS(root);

     printf("%d + %d = %d", n1, n2, n1 + n2);

     cin >> N;

     return ;

 }

总结：

1、注意最下面两层，和叶节点+叶节点上一层节点个数是不一样的。

巴特西

A1115. Counting Nodes in a BST

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