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1115. Counting Nodes in a BST (30)

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

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题解

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=1010;const int INF=1000000;struct Node{    int left,right;    int data;    int level;    Node(){        left=right=INF;    }}node[maxn];int MAXLevel=-1;void insertBST(int root,int key,int id){    if (key<=node[root].data)    {        if (node[root].left!=INF)        {            insertBST(node[root].left,key,id);        }else{            node[id].level=node[root].level+1;            if (node[id].level>MAXLevel)            {                MAXLevel=node[id].level;            }            node[root].left=id;        }    }else{        if (node[root].right!=INF)        {            insertBST(node[root].right,key,id);        }else{            node[id].level=node[root].level+1;            if (node[id].level>MAXLevel)            {                MAXLevel=node[id].level;            }            node[root].right=id;        }    }}int main(){    int n;    scanf("%d",&n);    int key;    for (int i=0;i<n;i++)    {        scanf("%d",&key);        node[i].data=key;        if (i==0)        {            node[i].level=1;        }else{            insertBST(0,key,i);        }    }    int firstAns=0,secondAns=0;    for (int i=0;i<n;i++)    {        if (MAXLevel==node[i].level)        {            firstAns++;        }else if (MAXLevel-1==node[i].level)        {            secondAns++;        }    }    //特判n==1    if (n==1)    {        printf("1 + 0 = 1");    }else{        printf("%d + %d = %d",firstAns,secondAns,firstAns+secondAns);    }    while(1){        getchar();    }    return 0;}

注：
1.题解采用静态构建BST的方法，减少了不必要的BFS过程
2.注意特判n==1否则第五个用例失败