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.
Each input file contains one test case. For each case, the first line gives a positive integer N (≤) 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 − will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.
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.
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
58 25 82 11 38 67 45 73 42
1 /* 2 Data: 2019-06-26 16:22:18 3 Problem: PAT_A1099#Build A Binary Search Tree 4 AC: 21:23 5 6 题目大意： 7 BST定义：lchild < root <= rchild 8 给定一棵BST的树形，将所给序列填入BST中，并打印层次遍历 9 输入： 10 第一行给出，结点数N<=100 11 接下来N行，给出第i号结点，左孩子的序号，右孩子的序号（0~N-1，-1为空，0为根） 12 最后一行，N个数 13 输出： 14 层次遍历 15 16 基本思路： 17 建立静态树，构建根结点与左右孩子之间的关系 18 中序遍历静态树，由于BST的中序遍历就是从小到大递增的序列，把排序好的序列依次填入树的结点中 19 层次遍历并打印输出 20 */ 21 #include<cstdio> 22 #include<queue> 23 #include<vector> 24 #include<algorithm> 25 using namespace std; 26 const int M=110; 27 priority_queue<int,vector<int>,greater<int> > bst; 28 struct node 29 { 30 int data; 31 int lchild,rchild; 32 }tree[M]; 33 34 void InOrder(int root) 35 { 36 if(root == -1) 37 return; 38 InOrder(tree[root].lchild); 39 tree[root].data = bst.top(); 40 bst.pop(); 41 InOrder(tree[root].rchild); 42 } 43 44 void Travel(int root, int len) 45 { 46 queue<int> q; 47 q.push(root); 48 while(!q.empty()) 49 { 50 root = q.front(); 51 q.pop(); 52 printf("%d%c", tree[root].data, --len==0?‘\n‘:‘ ‘); 53 if(tree[root].lchild != -1) 54 q.push(tree[root].lchild); 55 if(tree[root].rchild != -1) 56 q.push(tree[root].rchild); 57 } 58 } 59 60 int main() 61 { 62 #ifdef ONLINE_JUDGE 63 #else 64 freopen("Test.txt", "r", stdin); 65 #endif // ONLINE_JUDGE 66 67 int n,x; 68 scanf("%d", &n); 69 for(int i=0; i<n; i++) 70 scanf("%d %d", &tree[i].lchild,&tree[i].rchild); 71 for(int i=0; i<n; i++) 72 { 73 scanf("%d", &x); 74 bst.push(x); 75 } 76 InOrder(0); 77 Travel(0,n); 78 79 return 0; 80 }
PAT_A1099#Build A Binary Search Tree
原文：https://www.cnblogs.com/blue-lin/p/11090635.html
