pta 1099. Build A Binary Search Tree (30)

1099. Build A Binary Search Tree (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

这题信息量很大要用静态数组建树最后用层次遍历

#include
#include
using namespace std;
struct node
{
    int val, l, r;
}g[110];
int a[110];
queueq;
int n, k;
void mid(int k1);
void travel();


int main()
{


    scanf("%d", &n);
    for(int i=0;i     {
        int x, y;
        scanf("%d %d", &x, &y);
        g[i].l=x;
        g[i].r=y;
    }
    for(int i=0;i     {
        scanf("%d", &a[i]);
    }
    sort(a,a+n);
    k=0;
    mid(0);
    travel();
    printf("\n");
    return 0;
}






void mid(int k1)


{
    if(g[k1].l!=-1)
    {
        mid(g[k1].l);
    }
    g[k1].val=a[k];
    k++;
    if(g[k1].r!=-1)
    {
        mid(g[k1].r);
    }
    return ;
}


void travel()
{
    if(g[0].val==-1)
    {
        return ;
    }
    else
    {
        struct node e=g[0];
        q.push(e);
        int flag=0;
        while(!q.empty())
        {
            e=q.front();
            q.pop();
            if(flag==0)
            {
                printf("%d",e.val);
                flag=1;
            }
            else
            {
                printf(" %d",e.val);
            }
            if(e.l!=-1)
            {
                q.push(g[e.l]);
            }
            if(e.r!=-1)
            {
                q.push(g[e.r]);
            }
        }
    }
    return ;
}