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
题意:给你一棵二叉树的结构(n个结点),然后给你n个数,把树填进二叉树中,构成一棵二叉排序树。
思路:二叉排序树的中序遍历为递增序列,把二叉树结构建好后,输出中序遍历节点下标,然后再把给定的n个数递增排序。用一个数组来记录下标(0-N-1)与数的对应关系,然后再把二叉树结构层序遍历,并按下标顺序输出对应的数,既为结果。
#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
struct node
{
int data;
node * lchild;
node * rchild;
};
int n;
vector<vector<int> > vec;
vector<int> in_order_num(n);
vector<int> in_num(n);
node * creat_tree(int start){
if(start == -1){
return NULL;
}
node * root = new node();
root -> data = start;
root -> lchild = creat_tree(vec[start][0]);
root -> rchild = creat_tree(vec[start][1]);
return root;
}
int q = 0;
void in_order(node * t){
if(t == NULL){
return;
}
in_order(t -> lchild);
in_order_num[t -> data] = in_num[q];
q++;
in_order(t -> rchild);
}
void level_order(node *t){
queue<node *> que;
que.push(t);
while (!que.empty())
{
printf("%s%d",que.front() == t ? "" : " ", in_order_num[que.front() -> data]);
if(que.front() -> lchild != NULL){
que.push(que.front() -> lchild);
}
if(que.front() -> rchild != NULL){
que.push(que.front() -> rchild);
}
que.pop();
}
}
int main(){
cin >> n;
vec.resize(n);
in_order_num.resize(n);
in_num.resize(n);
for(int i = 0; i < n; i++){
int l, r;
scanf("%d %d", &l, &r);
vec[i].push_back(l);
vec[i].push_back(r);
}
for(int i = 0; i < n; i++){
scanf("%d", &in_num[i]);
}
sort(in_num.begin(), in_num.end());
node * tree = creat_tree(0);
in_order(tree);
level_order(tree);
return 0;
}