PAT甲级1115 DFS和BST

题目

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

题解

#include<iostream>
#include<vector>
using namespace std;
struct node{int v;struct node *left,*right;
};
node* build(node *root,int v){if(root==NULL){root=new node();root->v=v;root->left=root->right=NULL;}else if(v<=root->v){root->left=build(root->left,v);}else{root->right=build(root->right,v);}return root;
}
vector<int> num(1000);
int maxdepth=-1;
void dfs(node *root,int depth){if(root==NULL){maxdepth=max(maxdepth,depth);return;}num[depth]++;dfs(root->left,depth+1);dfs(root->right,depth+1);
}
int main(){int n,m;scanf("%d",&n)  ;node *root=NULL;for(int i=0;i<n;i++){scanf("%d",&m);root=build(root,m);}dfs(root,0);printf("%d + %d = %d",num[maxdepth-1],num[maxdepth-2],num[maxdepth-1] + num[maxdepth-2]);//等号前也有空格，格式要重视return 0;
}

回顾指针这个概念，它是存储着某变量的地址。

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PAT甲级1115 DFS和BST

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