- The left subtree of a node contains only nodes with keys less than the node's key.
- The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
- Both the left and right subtrees must also be binary search trees.
A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.
Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.
Sample Input:
10 1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4
这道题十分巧妙,当BST是完全二叉树时,它的数组存储即满足层序遍历。
思路为对数组表示的二叉树进行中序遍历,并按从小到大的顺序填入数组,最后顺序输出即为层序遍历。
#include<cstdio> #include<algorithm> using namespace std; int n,index=0; int a[1010]; int b[1010]; void inorder(int x){ if(x>n) return; inorder(2*x); b[x]=a[index++]; inorder(2*x+1); } int main(){ scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&a[i]); sort(a,a+n); inorder(1); for(int i=1;i<=n;i++){ printf("%d",b[i]); if(i<n) printf(" "); } return 0; }
最后附上我之前自己写的递归,代码存在死循环问题,但vs过期了,没有调试排错,留到以后解决
#include<cstdio> #include<algorithm> #include<queue> #include<cmath> using namespace std; int a[1010]; int b[15]; int n; struct node{ int data; node *left,*right; }*Node; node* change(int num,int l,int r){ sort(a+l,a+r+1); if(num==0) return NULL; node *root=new node; int t=0,now=num-1; if(now==0){ root->data=a[l]; root->left=root->right=NULL; return root; } while(now>2*b[t]) t++; if(now>=b[t]+b[t-1]){//zuoman root->data=a[l+b[t]]; root->left=change(b[t],l,l+b[t]-1); root->right=change(num-b[t]-1,l+b[t]+1,r); printf("%d ",root->data); return root; }else{//zuoweiman root->data=a[r-b[t-1]]; root->left=change(num-b[t-1]-1,l,r-b[t-1]-1); root->right=change(b[t-1],r-b[t-1]+1,r); return root; } } void bfsprint(node *r){ queue<node*> q; q.push(r); int tt=0; while(!q.empty()){ node *f=q.front(); q.pop(); printf("%d",f->data); if(++tt<n) printf(" "); if(r->left!=NULL) q.push(r->left); if(r->right!=NULL) q.push(r->right); } } int main(){ scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=0;i<=15;i++) b[i]=pow(2,i)-1; Node=change(n,1,n); //bfsprint(Node); return 0; }