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:

91 62 3-1 -1-1 45 -1-1 -17 -1-1 8-1 -173 45 11 58 82 25 67 38 42

Sample Output:

58 25 82 11 38 67 45 73 42

二叉搜索树 中序 排列一定是升序的。因此对数组排序然后中序遍历即可。

#include<stdio.h>#include<queue>#include<algorithm>using namespace std;struct node{int x;int left;int right;}stu[110];int cou=0;void inorder(int root,int a[]){if(root!=-1){inorder(stu[root].left,a);stu[root].x=a[cou++];inorder(stu[root].right,a);}}int main(){int n,i;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d %d",&stu[i].left,&stu[i].right);}int a[n];for(i=0;i<n;i++){scanf("%d",&a[i]);}sort(a,a+n);inorder(0,a);queue<int>q;q.push(0);int flag=0;while(!q.empty()){int head=q.front();q.pop();if(flag==0){printf("%d",stu[head].x);flag=1;}else{printf(" %d",stu[head].x);}if(stu[head].left!=-1){q.push(stu[head].left);}if(stu[head].right!=-1){q.push(stu[head].right);}}}