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
1 #include<stdio.h> 2 #include<math.h> 3 #include<set> 4 #include<algorithm> 5 #include<vector> 6 #include<queue> 7 using namespace std; 8 9 struct node 10 { 11 int l,r,v; 12 }; 13 14 node Tree[110]; 15 vector<int> vv; 16 int cnt = 0; 17 void inOder(int root) 18 { 19 if(Tree[root].l != -1) 20 inOder(Tree[root].l); 21 Tree[root].v = vv[cnt++]; 22 if(Tree[root].r != -1) 23 inOder(Tree[root].r); 24 } 25 26 int main() 27 { 28 int n,tem; 29 scanf("%d",&n); 30 for(int i = 0 ;i < n;++i) 31 { 32 scanf("%d%d",&Tree[i].l,&Tree[i].r); 33 } 34 35 for(int i = 0 ;i < n;++i) 36 { 37 scanf("%d",&tem); 38 vv.push_back(tem); 39 } 40 sort(vv.begin(),vv.end()); 41 inOder(0); 42 queue<node> qq; 43 qq.push(Tree[0]); 44 bool fir = 1; 45 while(!qq.empty()) 46 { 47 node ntem = qq.front(); 48 qq.pop(); 49 if(fir) 50 { 51 fir = 0; 52 printf("%d",ntem.v); 53 } 54 else 55 { 56 printf(" %d",ntem.v); 57 } 58 if(ntem.l != -1) 59 qq.push(Tree[ntem.l]); 60 if(ntem.r != -1) 61 qq.push(Tree[ntem.r]); 62 } 63 printf(" "); 64 return 0; 65 }