A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.
Each input file contains one test case. For each case, the first line gives a positive integer N (≤100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format left_index right_index
, provided that the nodes are numbered from 0 to N−1, and 0 is always the root. If one child is missing, then −1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.
For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.
9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42
58 25 82 11 38 67 45 73 42
解题思路:二叉搜索树中,给定序列排序后是中序遍历,树可以用一个数组表示,结构体存储每个左右儿子的信息,最后使用队列进行层序遍历
以下是代码:
//二叉搜索树排序后为中序遍历
//树可以用一个数组表示
#include
#include
#include
using namespace std;
const int N = 110;
struct Node
{
int left;
int right;
}a[N];
int n;
int w[N];
int tree[N];
int k = 0;
void dfs(int u)
{
if(u == -1) return ;
if(a[u].left != -1) dfs(a[u].left);
tree[u] = w[k ++];
if(a[u].right != -1) dfs(a[u].right);
}
void bfs()
{
queueq;
q.push(0);
vectorres;
while(!q.empty())
{
int t = q.front();
q.pop();
res.push_back(t);
if(a[t].left != -1) q.push(a[t].left);
if(a[t].right != -1) q.push(a[t].right);
}
for(int i = 0;i < n;i ++)
{
if(i) cout<<" ";
cout<>n;
for(int i = 0;i < n;i ++) cin>>a[i].left>>a[i].right;
for(int i = 0;i < n;i ++) cin>>w[i];
sort(w , w + n);
dfs(0);
bfs();
return 0;
}