solution Of Pat 1115. Counting Nodes in a BST (30)

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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结题思路：
题意要求我们对构建的树，分别计算最后两层的节点个数。
要求1：考察二叉搜索树的建立过程；
要求2：比跟节点大的插入到右子树，小于等于根节点的插入到左子树。

程序步骤：
第一步：建立二叉搜索树；
第二步：由于每个插入的新节点都位于叶子节点，故我们可以利用空间换取时间的思想，为每个节点增设depth属性，当前插入节点的depth=cur.parent.depth+1，由于当前插入为叶子，故其父亲节点depth必不为空；
第三步：遍历二叉搜索树所存储的数组，分别统计最后两层的节点个数。

具体程序(AC)如下：

#include <iostream>#include <vector>#include <algorithm>using namespace std;struct node{    int depth;    int value;    int left;    int right;    node():depth(0),value(-2000),left(-1),right(-1){}};vector<node> tree;int no=1;int deepest=0;void pushIn2BTree(int root,int value){    if(value>tree[root].value)    {        if(tree[root].right==-1)        {            tree[no].value=value;            tree[no].depth=tree[root].depth+1;            if(tree[no].depth>deepest)                deepest=tree[no].depth;            tree[root].right=no;            no++;        }        else            pushInBTree(tree[root].right,value);    }    else    {        if(tree[root].left==-1)        {            tree[no].value=value;            tree[no].depth=tree[root].depth+1;            if(tree[no].depth>deepest)                deepest=tree[no].depth;            tree[root].left=no;            no++;        }        else            pushInBTree(tree[root].left,value);    }}int main(){    int n,value;    cin>>n;    tree.resize(n);    if(n==0)    {        cout<<"0 + 0 = 0"<<endl;        return 0;    }    else    {        cin>>tree[0].value;        for(int i=1;i<n;++i)        {            cin>>value;            pushIn2BTree(0,value);        }//建树完成        if(deepest==0)            cout<<"1 + 0 = 1"<<endl;        else        {            int last=0,beforeLast=0;            for(int i=0;i<n;++i)            {                if(tree[i].depth==deepest)                    ++last;                else if(tree[i].depth==deepest-1)                    ++beforeLast;            }            cout<<last<<" + "<<beforeLast<<" = "<<last+beforeLast<<endl;        }    }    return 0;}