1115 Counting Nodes in a BST (30)
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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
题目大意:给出一串数,建立二叉树。找出最底层和次底层节点个数
思路:dfs()遍历树,每遍历依次某一深度的节点树加一;遍历到节点为空指针为止。还需要维护一个maxdepth变量,记录树的最深深度
#include<iostream> #include<vector> using namespace std; vector<int> ans(1001, 0); int maxdepth=-1; class node{ public: int val; node *left, *right; node(){left=right=NULL;} }; node* insert(node* root, int val){ if(root==NULL){ node* newnode = new node(); newnode->val = val; return newnode; } if(val<=root->val) root->left=insert(root->left, val); else root->right=insert(root->right, val); return root; } void dfs(node* root, int depth){ ans[depth]++; if(maxdepth<depth) maxdepth = depth; if(root->left!=NULL)dfs(root->left, depth+1); if(root->right!=NULL)dfs(root->right, depth+1); } int main(){ int n, i, t; node* root=NULL; cin>>n; for(i=0; i<n; i++){ cin>>t; root=insert(root, t); } dfs(root, 0); printf("%d + %d = %d", ans[maxdepth], ans[maxdepth-1], ans[maxdepth-1]+ans[maxdepth]); return 0; }
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