1099 Build A Binary Search Tree 甲级 xp_xht123

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.

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.

Input Specification:

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.

Output Specification:

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.

Sample Input:

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

Sample Output:

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;
}