1064 Complete Binary Search Tree (30 分) 建立完全二叉搜索树并输出其层次遍历

拼搏现实的明天。 2021-10-29 11:31 192阅读 0赞

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

给出n个结点，建立完全二叉搜索树并输出其层次遍历。

新做法~

#include <stdio.h>
    #include <string.h>
    #include <string>
    #include <map>
    #include <vector>
    #include <queue>
    #include <math.h>
    #include <algorithm>
    #include <iostream>
    #define INF 0x3f3f3f3f
    using namespace std;
    int n,p,l[1005],t[1005];
    void intravel(int index)
    {
    	if(index>=n)return;
    	intravel(2*index+1);
    	l[index]=t[p++];
    	intravel(2*index+2);
    }
    int main()
    {
    	int i;
    	scanf("%d",&n);
    	for(i=0;i<n;i++)
    	scanf("%d",&t[i]);
    	sort(t,t+n);
    	p=0;
    	intravel(0);
    	for(i=0;i<n;i++)
    	{ 
    		if(i) printf(" ");
    		printf("%d",l[i]);
    	} 
    }

这是我的老实人做法555

思路：构建一棵完全二叉的空树，然后把给定结点sort一下，中序遍历插入即可得。

与1099类似：[https://blog.csdn.net/TheWise\_lzy/article/details/99613082][https_blog.csdn.net_TheWise_lzy_article_details_99613082]

#include <stdio.h>
    #include <string.h>
    #include <string>
    #include <map>
    #include <queue>
    #include <math.h>
    #include <algorithm>
    #include <iostream>
    #define INF 0x3f3f3f3f
    using namespace std;
    struct Node
    {
    	int val;
    	int l,r;
    }tree[1005];
    void init(int n)
    {
    	for(int i=0;i<n;i++)
    	{
    		if(2*i+1<n)
    		tree[i].l=2*i+1;
    		else
    		tree[i].l=-1;
    		if(2*i+2<n)
    		tree[i].r=2*i+2;
    		else
    		tree[i].r=-1;
    	}
    }
    int a[1005],p;
    void Inorder(int x)
    {//中序遍历建立BST 
    	if(tree[x].l!=-1) Inorder(tree[x].l);
    	tree[x].val=a[p++];
    	if(tree[x].r!=-1) Inorder(tree[x].r);
    }	
    int main()
    {
    	int n,i,j;
    	scanf("%d",&n);
    	init(n);
    	for(i=0;i<n;i++)
    	scanf("%d",&a[i]);
    	sort(a,a+n);
    	p=0;
    	Inorder(0);
    	//层次遍历 
    	vector<int> v;
    	queue<Node> q;
    	q.push(tree[0]);
    	while(!q.empty())
     	{
     		Node u=q.front();
     		q.pop();
     		v.push_back(u.val);
     		if(u.l!=-1)
     		q.push(tree[u.l]);
     		if(u.r!=-1)
     		q.push(tree[u.r]);
     	}
     	for(i=0;i<v.size()-1;i++)
     	printf("%d ",v[i]);
     	printf("%d\n",v[v.size()-1]);
    }

[https_blog.csdn.net_TheWise_lzy_article_details_99613082]: https://blog.csdn.net/TheWise_lzy/article/details/99613082