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
#include<cstdio> #include<queue> #include<algorithm> using namespace std; const int maxn = 110; struct Node{ int data; int lchild,rchild; }node[maxn]; int n,in[maxn],num = 0; void inorder(int root){ if(root == -1) return; inorder(node[root].lchild); node[root].data = in[num++]; inorder(node[root].rchild); } void BFS(int root){ queue<int> Q; Q.push(root); int num = 0; while(!Q.empty()){ int Now = Q.front(); Q.pop(); printf("%d",node[Now].data); num++; if(num < n) printf(" "); if(node[Now].lchild != -1) Q.push(node[Now].lchild); if(node[Now].rchild != -1) Q.push(node[Now].rchild); } } int main(){ int lchild,rchild; scanf("%d",&n); for(int i = 0; i < n; i++){ scanf("%d%d",&lchild,&rchild); node[i].lchild = lchild; node[i].rchild = rchild; } for(int i = 0; i < n; i++){ scanf("%d",&in[i]); } sort(in,in+n); inorder(0); BFS(0); return 0; }