1099 Build A Binary Search Tree (30分)

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.

figBST.jpg
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

#include<bits/stdc++.h>

using namespace std;

const int N=110;

int n;

int l[N],r[N],v[N];
int inorder[N]; 

void dfs(int u,int &k)  //整个dfs使用一个k,加引用，也可直接定义成全局变量
{
    if(u==-1) return;
    dfs(l[u],k);
    v[u]=inorder[k++]; 
    dfs(r[u],k);
}

void bfs(int root)
{
    queue<int> q;
    q.push(root);

    bool flag=true;
    while(!q.empty())
    {
        auto t=q.front();
        q.pop();

        if(flag)
        {
            cout<<v[t];
            flag=0;
        }
        else cout<<' '<<v[t];

        if(l[t]!=-1) q.push(l[t]);
        if(r[t]!=-1) q.push(r[t]);
    }
}

int main()
{
    memset(l,-1,sizeof l);
    memset(r,-1,sizeof r);

    cin>>n;
    for(int i=0;i<n;i++)
    {
        int a,b;
        cin>>a>>b;
        l[i]=a,r[i]=b;
    }

    for(int i=0;i<n;i++)
        cin>>inorder[i];

    sort(inorder,inorder+n);

    int k=0;
    dfs(0,k);

    bfs(0);

    return 0;
}