pat 甲 1099. Build A Binary Search Tree (二叉搜索树)

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

tips:先建立一棵树，然后数组排序后按照中序遍历的顺序逐一插入

#include<iostream>#include<cstring>#include<queue>#include<algorithm>using namespace std;int t[110][2];int a[110];int b[110];int n,sz;void level(){queue<int>q;q.push(0);while(!q.empty()){int tt=q.front();q.pop();if(!tt)cout<<b[0];else cout<<" "<<b[tt];if(t[tt][0]!=-1)q.push(t[tt][0]);if(t[tt][1]!=-1)q.push(t[tt][1]); }}void dfs(int x){if(t[x][0]!=-1)dfs(t[x][0]);b[x]=a[sz++];if(t[x][1]!=-1)dfs(t[x][1]); }int main(){cin>>n;for(int i=0;i<n;i++)cin>>t[i][0]>>t[i][1];for(int i=0;i<n;i++)cin>>a[i];sort(a,a+n);dfs(0);level();return 0;}