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 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
給定條件:
1.給定n個節點
2.這n個節點是屬于一顆二叉搜尋樹的節點,并且這棵樹是完全二叉樹
要求:
1.這棵樹的層次周遊結果
求解:
1.因為是二叉搜尋樹,是以将n個節點排序後可得到該二叉樹中序周遊的結果
2.如果根節點的編号為x,那麼它的左子樹根節點是2*x,右子樹根節點是2*x+1 (x!=0),編号順序即層次周遊順序
3.根據中序周遊結果,更新層次周遊中的節點資訊
#include<cstdio>
#include<algorithm>
using namespace std;
int n, id;
int a[1005];
int ans[1005];
void dfs(int index){
if(index > n) return ;
dfs(index*2);
ans[index] = a[id++];
dfs(index*2+1);
}
int main(){
while(scanf("%d",&n) != EOF){
id = 1;
for(int i = 1; i <= n; i++){
scanf("%d",&a[i]);
}
sort(a+1,a+n+1);
dfs(1);
for(int i = 1; i <= n; i++){
if(i!=1) printf(" ");
printf("%d", ans[i]);
}
}
return 0;
}