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.
A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.
Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.
Sample Input:
10
1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4题意：
给出一组数字，用这组数字构成一棵完全二叉搜索树。
思路：
完全二叉搜索树中序遍历的结果，就是升序排序的结果。所以我们可以根据结点的个数，构造出一棵完全二叉搜索树的框架，然后再中序遍历这棵树，将所给的数组排序后依次插入即可。
Code：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 typedef struct Node* node; 6 7 struct Node { 8 int val; 9 node left; 10 node right; 11 Node(int v) { 12 val = v; 13 left = NULL; 14 right = NULL; 15 } 16 }; 17 18 void buildTree(node& root, int num) { 19 queue<node> que; 20 que.push(root); 21 int count = 1; 22 while (count < num) { 23 node temp = que.front(); 24 que.pop(); 25 temp->left = new Node(-1); 26 que.push(temp->left); 27 count++; 28 if (count < num) { 29 temp->right = new Node(-1); 30 que.push(temp->right); 31 count++; 32 } 33 } 34 }; 35 36 int tempIndex = 0; 37 void inorderTraveral(node& root, vector<int>& keys) { 38 if (root == NULL) return; 39 inorderTraveral(root->left, keys); 40 root->val = keys[tempIndex++]; 41 inorderTraveral(root->right, keys); 42 } 43 44 void levelTraveral(node& root) { 45 queue<node> que; 46 que.push(root); 47 bool isFirst = true; 48 while (!que.empty()) { 49 node temp = que.front(); 50 que.pop(); 51 if (isFirst) { 52 cout << temp->val; 53 isFirst = false; 54 } else { 55 cout << " " << temp->val; 56 } 57 if (temp->left != NULL) que.push(temp->left); 58 if (temp->right != NULL) que.push(temp->right); 59 } 60 } 61 62 int main() { 63 int n; 64 cin >> n; 65 vector<int> keys(n); 66 for (int i = 0; i < n; ++i) cin >> keys[i]; 67 sort(keys.begin(), keys.end()); 68 node root = new Node(-1); 69 buildTree(root, n); 70 inorderTraveral(root, keys); 71 levelTraveral(root); 72 return 0; 73 }